C.S.

III RDVANCED ENGINEERING MATHEMATICS

RANDOM VARIABLES
1
PREVIoLS YEARSQUESTIONs
For a continuous uniform distribution over an interva! la, bË:
Mean (p)
PART-A a +b

Variance (o)
Q.1)What is difference between skewness and kurtosis. o-(6-a)²
RTU 2024]
12

Ans. Difference between Skewness and Kurtosis:

Standard Deviation (o)
Skewness Kurtosis (b-a b-a
The characteristic of a Kurtosis means the 12 /12
frequency distribution relative pointedness of
the standard bell curve, 2. Exponential Distribution : The exponential
that. ascertains its
distribution is commonly used to model the time between
symmetry about the mean dfined by the frequency evnts in a Poisson process, where events occur
-distribution.
is called skewness.
measure of continuously and independently at a constant average rate.
Skewness is a measureof Kurtosis is a
degree of taildness in the For an exponential distribution with rate parameter >0:
the degree of
lopsidedness in the frequency distribution. Mean ()
frequency.istribution.
Skewness is an indicator Kurtois is a measure of ".

data, that is either peaked

of lack ofsymmetry, ie. or flat, with respect to the
Variance ()
both left and right sides probability distribution. 1
of the curve are unequal,
with respect to the central!
point. Standard Deviation (o)

Q.2)What is mean, variance and standard deviation of

distribution.
form distribution und exponential |RTU 2024)
OR DTind the value of theconstant c such th¡t the
(RTU 2019]
Define he exponential distribution. cv, 0<x<3
Otherwise is aprobability density
uniforim distributionis
Ans. Uniform Distribution :Thealloutcomes are equally function. (RTU 2023)
a probabis distribution where
interval..
Jikely within a defined
 B.Tech. (I Sem.) C.S. Solved Papers
AEM.2
.Ans. f(*)=cx, (0<x<3) QIfE) =2 and E() $, then what is the value of
(RTU 2022]
: For probability density function E(2X+3 Y)?

Ans. E(2x + 3y) 2E(x) +3E()

=2 x 2+3 x 5
4+15
19
=1
0.8 What is the coefficient of skewness, if the mean
and mode of the distribution are equal? [RTU 2019)

Ans. If the mean and mode of the distribution are equal,

then the distribution has zero skewness.

A4 If E)=4 and E() =1, then what is. thevalue of

E(2X + 3)? RTU 20231
Mean Median =Mode
Ans. EX) =4 Fig.
E(Y) =1, Q.9hat is the variance of thePoisson distribution with
E(2X +3Y)= 2E(X)+3E(Y) mean value 5? -{RTU 2019)
=(2x4) +(3*1)=8+3=11
Ans. In' the Poisson distribution
Q.5)Write Chebyshev's inequality. (RTU 2022 mean = variance
So variance will be equal to S..
Ans. Chebysbev's Inequality : Chebyshev's inequality
is aprobability theory that guarantees that within aspecified 10 The probability density function of the random
range or distance from the mean for a large range ofvariable x is given by:
probability distributions, no more than aspecific fraction of
vaiues willpresent.
for 0<*<yfind the value of K.
f)={Vx0, elsewhere
qE Given the function f(x)= RTU 2018]
0,
Is this function a density function? (RTU2022] Ans. Probability density function is given by

Aos. We know thet if the function satisfy f(x) for0<x<y

0, elsewhere
:. Above giveh is a pdf, so
Then it is called pdf.

K
So the fx) is a pdf.
 AEM.3
Advanced Engineering Mathematics
Papers Let E be a random experiment and S be a sample
space associated with it, a function X(s), where seS. is
2 dx = 1 called a random variable.
value of
RTU 20221
X
s. -X(s)
K =1
1
2 Jo Fig.
Suppose that our experiment Econsists of tossing two
fair coins, etting Xdenote the number of heads appearing
ie mea then X isa random variable taking one of the values 0, 1,2
TU 2019] 1 with respective probabilities,
K=
re equal, 24/y| P{X-0}=P{(T))=
O.11 Find the correlation coefficient between z and y P{X=}=P{(H.T),(T, H)} -;
when it is given that.
n= 15, x= 50, Ey=-30, x= 290,
Ey = 300, Zey = I15 |RTU 20181
P{X-2}=P{(H,H) =4
Ans. The coefficient of correlation is given by We have P{X= 0} +P{X =1}+P{X=2}=1.
Hence, X is a random variable taking values 0, 1, 2. A
Cion with
xy-ù random variable is also called chance variate or stochastic
variate.
TU 2019]
Vn
.:.Given that n= 15, Ex= 50, Ey=-30, EX= 290, Zy'= PART-B
300, Exy=-115
X 50 10
15 3 QA3 Thejoint probability mass function of (X, Y) is given
random by:
Ey-30 =-2
n 15 i=,2j=1, 2,3
20 otherwise
ue of K.
15 x-115) () Find .
(i) Find the marginal probability mass function of x
RTU 2018
V15
-43
1003004
290-Nis and y. (RTU2024]

3 Ans. (1) Given,

I58 100 i-1,2;j=l,2,3
/16
V3 otherwise
r=04358 Ans.
By condition of probabilitymass furnction (PMF)
Q.1 What Is arandom variable?
i=l jl
Ans. Random Variable : A random variable X is a function
that associates a real nunber'with each element in the
sample space S. i=l j=l
 AEM.4 B.Tech. (M Sem.J C.S. Solved Papers

-i[l1+2]
Let, 4=1, x,=2
y =1, y, =2, y, =3 1 1

: a|P-1+-2+-3+2-1+2-2+2-3|=1 30

a[1+2+3+4+8+12]=1
A(30) =1 -aP2+2-2]
30
(ii) Marginal PMF ofx
1
-x10=.
30
j=l

,(3)-EPr(3)
P,0)-SP,(.y;)
j=l
-[3+23]
=

:2*3)
30 2
305
Q4The distribution fuiction for a random variable Xis
P,2)=}P,(2.y)
jl
P(a)=-e,x<0
-a2'.1+2".2+2'-3]. Find (a) the densiy function and (b) P3<*s4).
(RTU 2023)
l4+8+12] Ans. Given,

=-x24 F(:)-1-e*, x20

30 0. x<0
4 : We know that

Marginal PMF ofy () Density function f(x) =F()

Bo)-R.() =0+2e-2x
=2e-2x
0)-}P.(x,!)
ial
 Advanced Engineering Mathematics AEM.5
m.j C.S. Solved Papers
D.F.= 2e2*; x2 o Q.16 4 box contains a' wlite and 'b' black balls, c
balls are drawn. Find the expected value of the number.
6) P(-3<xs4) of white balls drawn. [RTU 2019}
: -3<xS4
{-2,-1,0,12,3,4} ex Ans. let the random variable Xdenote the event of drawing
but -2,-l are undefined values for Density Function. one white ball.
P(-3<xs4)=F(4)-F(0) Hence P(X=0)=P(1black ball is drawn)
b
a+b
Upper Lower
Limit Limit and P(X =l) = P(l white ball is drawn) = a+b
F(:)=(1-) b
F(4)=1-e1X) =]-e !
Hence E(X) =0.-a+b-+l.a+b a+b.
F{0) =i-e?00) =1-el=1-1=0 As C bails are drawn, hence the required expected value.
ac
P(-3<xs4)=i-e*-o ial a+b a+b
-(1-e)
Q.17 The joint probability density function of a
Q.15IfX is uniformly distrbuted with mean 1 and two-dimensional random variable (X, ) is given by :
variance 4/3, then estimate PX< 0), -[RTU 2023]
[2; 0<*<1,0<y<x
Ans. Given:xis uniformly distributedfunction with, |0;. elsewhere
Mean =1
Variance=4/3 Find the marginal density functions of X and Y. Also
fnd the conditional erisity function of Ygiven X =x
Ifxis uniformly distributed function, and conditional density function of X. RTU20191

Then, f(x)= (Where aand bare two variables)

Ans.(i) Marginal density function ofX and Y respectively'
a+b are

random yariable Xis

Its Mean
a +b =2 ..():
and Variance - 2dy =2(y), =2x; 0<x<I
12
(bj P(-3 <r<4).
b-a =t4 ..(i)
[RTU2023) the f9) =,f(xy)dk
On solving the equations (i) and (i) and finding and
values of a and b we get,
a=-1, b=3 =f2dx =2(x=2;0<y<x.
(i) Conditional density function ofY, given X=x
b-a 4
f(y) 2-=;0<x<1
1
f. (x) 2x -x
and conditionaldensity function of X, given Y=y.

f(x)2-1;0<y<x.
fy (y) 2
 B.Tech. (I Sem) C.S. Solved Pepers

(c) Similarly marginal probability distribution ofy is given

QA7he jon prodabiliy mass junction of (, ) is given by
) k( + x0, I, 2; y 1, 2,3. Find:

o)Atnrgnal probabitity distribution of X.

Marenol probadiiy distribution of Y. 15
fid anational probabiliy istribution of Xgiven y mI. Po +Py +P=1Sk =
72
|RTU 2018) 24
Po +Pit P =24k =
Ans. The Nmt prhability distribution of (x, y) can be
Nentdin tablular form as: 3 33
Po +Piy +Pa33k
72
3 (d) Conditional distribution of xgiven y=1is
9k
Sk Ik -(using(i) it can be
7 13k y=l P 15k
(a) As above given is a pmf, hence represented in tabular föm

P, 1Sk
Po 3k 1
ISk 15k 5
1
BL_Sk
1Sk 1Sk 3

P 7k 7
1Sk 1Sk 15

PART-C
-
24
09 Arandom variable x has the following probability
72 distribution:
01 2 3 6 7
36 2K' Tk+k
() Find k.
() Evalnate P (<6), P(> 6) and P(0
Hence Marginai probatbility distribution ofxis:
<I<5).
(i) Find distribution function of x.
(iv) FìndP 1.5 <r<4.5\
[RTU 2024]

Ans. (1) Given probability distribution

P(x)=0, k, 2k, 2k, 3k,k, 2k, 7k?+k (iv

P(1) -1
 AEM.7
Ders
Advanced Engineering Mathematics
Iven k+ 2k +2k +3k +k?+ 2k? + 7k' +k =1 P(x=3)+ P(x=4)
1-[P(x=0) +P(x =1) +P(x= 2)]
10k +9k-1=0 2k+3k
(10k-1)(X +1) =0 1-f0+k+2]
2,3
10 10
(i) P(x <6) = P(x= 0) + P(x=1) t... P(x=S) 1, 2
P(x<6) 0+k+ 2k +2k +3k +k?
- 8k + k?
I-*10 1o
10
3
10
=0.8+ 0.01
=0.81 10 5
P(x>6) 77
10
=1-P(x<6)
=]-0.8! Q.20 Joint distribution function of two discrete random
=0.19 variable Xand Yare given by f(x, y) = c(2x+y). Wlhere
x and yassumes all nteger values such tlhat 0 <xs2,
P(0<x<5) = P(x= 1). +P(x=2) +P(x =3)
+P(x=4) 0 Sy s3. Find :
=k+ 2k+2k +3k (i) PX= 2, Y=1)
(i) P(X 21,Y s2)
=8k =8x-L-0.8 (iv) Marginal distributions
10 () Check the dependency |RTU 2023, 221
(i) Distribution function
XF(x) = P(X<) Ans. ()
1. 2 3
1 k=1/10 0 2c 3c
2 3k=3/10 2c 3c 4c
3 Sk= 5/10 2 4c Sc 6c 7c
4 8k= 8/10
8k + k²= 80+] 81
10 100 100 100
42c =1
6 8k +31? =. 8-3 80+3 83
10 100 100 100
7 9k + 1029+
1o 90+10 100 42

10 100 100 100 (ii) P(X=2, Y=1)=f(2, l)= x 5 = 5

42 42
(ii) P(X>1,Y s 2) =f(1,0) + f(1,1) + f(1, 2) + f(2,0)
(iv) x<2 +f(2,1) + f(2,2)
2c+3c + 4c + 4c + Sc + 6c
P(1.5<x<45)n(>2)]
P(*>2)
=24c
4
24 x
42 7
 Mean-Qtb
2
AEM.8
B.Tech. (i Sen.) C.S. Soived PaperS
(iv) Marginal distribution ofX,
Ans. LetX denote the number of defective items. Hence
X=0,1,2,3, 4as four items are drawn. Defciive items =
5and nondefective items =20.
=f(x,0) +f(x,1) +f(x,2) f(x,3) () In thiscase probability distribution will be
P(X=x)
(2x+2x +1+2x +2+2x +3) =0.3830

4 (8x+6)
=0.4506

f(0)- 21 (4x +3);0<xs2 2 20

C =0.1502
Marginal distribution of y
3

f(y)= 2f(.y)=f0,y) +f(1,y)+{(2, y) =0.0158

4
=0.0004
n+2+y+4+y)
=
25c.
1 Hence E(X) =0x 0.3830 +1 x0.4506+2
f9)-3yt6) x0.1502 +3 x0.0158 +4 x0.0004
-0.8
,(y+3); 0sys3 () As here the items are replaced, hence the
being defective and non defective remains probability of
constant.
(v) Since
Hence P(defective) =0.2=p
25
20
147 147 98 49 P(non defective) 25 =0.8=q
Ix,
-+ ly =f(xy)
Hence probability distribution is :
21 42 X
So x and y are not independent. P(X=x)
*Cop'*=0.4096
0.21 From a lot of 25 items containing 5. defectives, a 2
"Cp' =0.4096
sample of4 items is drawn at random, "C,p'q =0.1I536
() without replacement (i) with replacement 3
Find the expected value of the number of defectives in 4
Cp'g' =0.0256
the sample in each case. JRTU 2018 *C,p'q' =0.0016
Hence E(X)=(0x0.4096) +(1 x 0.4096) +(2 x0.1536)
+(3 x0.0256) +(4 x0.0016)
=0.8

choace fumio B
(

Veianece
 Poisov. Diot
Binomiol Distbutio
wed Papers Meen) n meom Vaúan ce
ems. Hence
ive items =
Aleit)

BINOMLAL DISTRIBUTION 2
PREVIoUS YEARS UESTIONS
,

Q.2)What is spearinam rank correlation? |RTU 2024|

PART-A:
Ans: Spearman Rank Correlation: Spearman's rank
642 correlation measures the strength and direction of
4 x0.0004
Q.1 Fit a straight line of followingset ofobservation:
association between two anked variables. It basically gives
the measure of monotonicity of the relation between two
4 variables ie. how well the relationshíp betweèn two
obability of variables could be represented usingamonotoric function.
Stant. 2 68
RTU 2024]
The forinula for Spearman's rank coefficient is:
p=l- 624
Ans. Straight line equation n(n-1)
y= +b ..)
Normal equation 0.3Define Binomial distribution and wrüe its mean and
}y= ax+nb ..) Variance. .RTU 2023!
Zy=a+b: ..i) Ans. Binomial Distribution: The binomial distribution is
n=5 a discrete probability distribution that módeks the number
YV of successful outcomes in a fixed number of independent
trials, where each trial has only two possible outcomes:
2 4 success or failure. It is often used when there are set
3 6 9 18 number of trials and a constant probability of success in
4 16 32 cach trial.
2x0.1536) 10 25 50

40.0016) Ex=15, Ey =30, x =55, Exy =110 Mean : }=E(X) -r P(x =)

[=0
Puttingthevalues of Zy, x, S' and Exy in equation (i) n

and (iij -Zr"c,p'q

re0
30=a.15+ 5.b ..i)
110=a55 + b.15 ..(v) n
n!
Solve the equation (iv) and (v), we get
u=2, b=0
Putting the values of a and b in equation (i) n
y2x + 0 (n-1)!
y=2x
(-){(n-)-(r 4),P q )

Mean
 The normaldistributton, also
AEM.10 Ans. Normal Distribution: continuous probability
is a
known as the Gaussian distribution, mean, describing data
distribution that is symmetric about its
central value. It is one of the most
ral
that clusters around a
statistics and real-world
}-EX)= np(q +p)=np.: { ptq=1} widely used distributions in
phenomena.
Variance : E(X) =p(x=r)
n=0 chiüdren each, how
Q.6 Out of 800 families with four one
many families would be expected to have at least
Xrr-)+r) "C,p'q boy? Assume equal probability for boys and girls.
[RTU2018]

n!
=2rr-)Pa}r"C,p'q Ans. Let.random variable_x denote the number of boys.
1
Here n= 4, N=800, p=q= 2
-)-(r-2)P).
(n-2)! *E(X)
=n(n -}p 2r-2){(0 P(x=r) =C,p'qt
P(x 1) =P(x = 1)+ P(x =2) + P(x=3) + P(x=4)
= n(n-)p' Cpqo-)-(r-2) +np
'n=2

-=n(n-1p tnp
Sovariance(X) =E(X)- (E(X)Y
=n(n-1)p+np-np:
=np-np+np--np
i= np(-p)
=npq {"ptq=1}
15
16
(Q4HoN miany number of normalequations arerequired
jor fiting apolynomtal of mdegree, by least square 15
method? 1RTU20731 | Here number of families having atleast 1boy =800x 16
=750
Ans.(m+1) normal equations will be required for fitting a
polynomiaB of mdegree by least sqüare method. (Q.7Define Binomial distribution. /RTU 2018]
For example, y à+bx + Cx isapolynomial equation
with degree 2. Ans. Refer to 0.3.
Now, Ey÷ na+b£x +«x?
PART-B
Exy =ax+bEx? +ox
Exy= a~x'+b2*+¢x
.:

Hence we require three normalequations for fitting å Q.8 Caiculate the coefficient of correlation ana obtain
polynomial ofdegree 2: : lines ofregression for the following data. (RTU 2024, 19}
2 34 5 6 8 9
(Q.spefine normat distribütion: . [RTU 2022] Ly 10| 12 11 13 14I6 15
 Advancd EngineeringMathematics AEM.11)
Ans. Ans, Poisson Distribution: There exist situations where
the probability P of the happening of an event is very smal,
ta 1 81 but the number of trials n is very arge so that the event can
64 16 happen several times.
10 100 30
Example : Number of persons born blind per year in a
4 12 16 144 48 certain city.
25 121 55 Example :Number of accidents that take place on a busy
13 36 169 78
7
road at any particular instant.
14 49 196 98
ne
16 64 256 128
The probability ofgeting 0,1,2,3, ...Such event is
15 81 225 135 P(r) =emrwhere = 0,1,2, ...
Ly r!
Exy =
45 108 285 1356 597 Mean and Variance of Poisson Distribution:
S
Here n=9, m
Mean =p:x; =>re-m r!n (r-1)!
= x 45 T=0 rel

9 m? m²
m -+
=e
1!2 31
-2_108-=12 = me " e" =m
9
Mean = m

() Variance = p;x-(?=)ren"-m? r=0

|285 |60
-25 =V9 r!
r=0
G, -2.58
m
+ m-m2

ze"r-2)!
= m'em
m-z
-+m-m=m +m-m =m
-2)!
i5 1356 60
-144 =
16 V9 So, Variance =m, Standard Deviation Jm
9, = 2.58 Limiting Case of Binomial Distribution : Let x be a
binomial variate with parameters nand pthen
018 y-y and
P((=P(x=)C,P(1-p)*", r=0,1,2,..n
ptq=1
Let us make following conditions:
(0 Number of trials are indefinitely large i.e. n’ o
(i) Probability of success for each trial is very small
597 S.t, P ’0 and np mis finite.
-60
9
(2 58)<(2.58) So, P(r)=
n!

I095 rl(n
bsain put p=m/n if n’ othen p -0
limiting case
Define poisson distribution. Derive it a
4, 191
mean and variunce
ar hinomial istribution. Find the (RTU2024) so p(r)
n! m/n
15 also. rl(n -r)! 1
 5.Tecl. (T Sem} C.S. Solxd Papers
n(n -1).(n -r+l) m
Q.11 Derive moment generating functionfor Binomial
distribution. (RTU 2022)

m Aus. E(e)" -)'Cp'qe

-"C, (pe)q=(g+ pey
Take lim n ’ , we get Differentiating w.r:t. t' we get
M,()=n(g+pe pe'
lim P(r) = lim m'
- On putting t=0, we get , n(qtpp
r!
4 =np, M() =e*M(t)
Moment generating funçtion of the binomial distribution
m'e about it's mean m=np is given by
- r=0,1,2...co
r! M(t) =eM,t
Which is the probability distribution for poissiòn distribution M(te pq +pey
thus poission distribution is the limiting case of Binomial
distribution. =(qe+pe
-(qei+ pefha
Q.10 Discuss the rank correlation coefficient for the data
given below: (RTU 2023]
12 15 14 19
40 41 48 60 50
-p+2p,p
2 2! 4!

Ans.
3!
S.No. xi R Bi,
1. 10 5 40 5 0 Pqt +:
2! 2! 3!.
2. 12 4 41
,P'gp'gt +...
3. 5 2 48 3 -1 4! 4!
4, 4 3 60 1 2 4

19 1 50 2 -1 .

n=S and L4' =6

Rank =] 6E4'
Equating the coeficients of like powers of t on both sides,
We get
6x6 36
npq, M=npqlg-p).H,-npq[l+3(1-2)pa)
(5)(24)
--0.3 = 0.7 (upg)'
bl eo Jinas o gesiou.
Angle
 xdPapers Tepeed rank coSUlaio
Advanced Engineering Mathematics
nla1) 9--GEdE)
-r Binomial AEM.13)-byx
|RTU2022] npg 12 =np[q+p)l+(a-1)p(g+p=np[1 +(n-1)pl
=np[np+(1-p)]=np[np+t]=rtnpq
-Puting values in equation (1)
Vrpg
Coefficient of Kurtosis is given by
H_npgl +3pg(n-2)) =3+|6pg S.D.= g=npg
1 =npq |5y=sx
(npg) npq Hence for bionomial distribution

Y2=By-3= 1-6pg Mean =np

npq 4onpg (variance)
Mean:
Q.12 Fit a straight line to the followingdata regarding
distribution
x as independent ariable:
..+"C,
EP,-nqp+n(o-1)qp 1.0
2 3 4
1.8 3.3 4.5 6.3
2 (RTU 2022]

Àns. Suppose a straight line to be fitted to the given data is

a under,
y=a+bx .. (1)
=(q+p)°=4 Then the normal equations are
Since qtp=1
=na +b)x ...(2)
Mean = <np
Lxy -a)x +b* .. (3)
Standard Deviation: Now from the given data

. (1)
11.8 1.3
3! ris the deviation of items (success) fromequation (1) |23.3 6.6
Ef=1, 2fr -np 34.5 13.5 9
416.3|25.2 16
3n(n-1)(n-2)
Eff=0nqp +2n(a-1)q+ 2 Now substituting all, there values in the normal
equations, we get
16.9 = Sa+ 10b . (4)
47.1 =10a +30b ...(5)
3!
3Án-IXn-2),
2!
Solvang equation (4) and (5)
a=0.72, b=1.33
Thus the required equation of straight line is
n-1)gp+ 2
y=0.72+1.33 x
both sides,
pg] *nplq*(n-g*pt (n-1X7-2) Q.13 Fit a straight line to the following data :
1 2 3 4 68
2.4 3 3.6 - 4 5 6
+(a-1)plg+{n-2)4pt (n-2(1-3)
2! (RTU 2019]

Yat b
(AEM.14 B.Tech. (M Sem.) C.S. Solved Papers
Ans. Let the line to be fitted is Also from figure
y=a+ bx .. (1) P(65 <x<)=0.5-(0.40 +0.05) =0.05
By the principle of least square method, we get the As z
normal equations as :
}y =6a+ bx ".. (2)
60-u
and Exy =a x+ bEx ...(3) Hence for x=60, we have -Z = ..()

2.4 1 2.4 65-p

And for x -65, we have -z, ...()
3 4 6
3 3.6 10.8 As P(x<60) =P(z<- z) =0.5 -P-z, <z<0)
4 16
4 16 =0.5 - P(0<z<z)=0.05
6 36 30 =P(0<z<z)=0.5-0.05= 0,45
6 64 48
}x=24 }y=24 Using normal table, we have z = 1.64
Ex=130 }xy= 113.2 Also P(65 <x<)0.05
Substituting these values in equation (2) and (3), we get
24 = 6a + 24b...(4) ’ P(-z<z<0)= 0.05
013.2 =24a + 130b ..(5)
P(0<z< z,)=0.05
Due to symmetry]
Solving equation (4) and (5), we get Using normal table z, =0.13.
34b =17.2’b=0.506 ..(6)
Substituting it in the equation (1), we have
Substitutingthese value in equation () and (ii), we
get
6a =24-(24 x0.506)=11.856
60 3-4
11.856 - 1.64 and - 0,13
-=1.976
Dividing them we get
Hence the best fitted straight line is 1.64
y=1.976+0.506x 60
0.13 65-4
Q.14 Of a large group of men S% are under 60 inches (1.64 0.13) =(65 x 1.64)- (60 x0.13)
(1.51) =98.8
in height and 40% are between 60 and 65 inches.
Assuming anormal distribution find the mean height 98.8.=65.43
and standard deviation. RTU 2018] 1.51
Also from equation ()
Ans. Letx denote the normal variable denoting the height (65-65.43)
ofa large group of menwith mean and standard deviation =

0.13
o.

Given P(x <60) =0.0S 0.43

-=3.30
And P(60 <x<65)=0.40. Hencex=60 and x= 0.13
65 should lie on left hand side of u.
Q.15Define rectangular distribiution andfnd its inean
and vaiance. RTU 2018]
0.40 Ans. Rectangular Distribution : Acontinúous random
.05
variablex is said to follow a continuous rectangular
distribution over an interval (a, b) if its p.d.f. is given by:
Jk ; a<x<b|
X=60 x*65 0 ; otherwise
z=0 Where k is a constant.
Fig. As aböve is a p.d.f. hence
 Advanced Engineering Mathematics AEM.15
a' +b'-2ab
ff(x)dx =1 12

(b-a
kdx =1. 12

Hence méan =
a+b
variance=:(B-a)
k[b - aj =1 2 12
(i) 1
k
0) b-a 0.16Ina hormal distribution, 31% of the itens are under
Hence the p.d.f. of xis given as 45 and 8% are over 64. Find the mean and S..D, of the.
1 Z
distribution. Given that .dt, the values
f(x) =b-a
otherwise of Zcorrespönding top'= 0.19 and p 0.42 are 0.50
here x is known as uniform variate with parameter a and 1.40 respectively. (RTU 2017
and b.
Mean and Varianc of Rectangular Distribution : ByAns. etus consider Hand oas mean ahdstandard deviation
we defi. of moments about originare given as ofthe distribution
P(<z<0)=P(0<z<a)=0.19

0.19042
b-alr+i) 31 .08
1
b-a r+1 Given
-X=45 X=# X=64
In particular
r= P(*<4s)=031
and. P(X>64)=0.08:
H=Mean = 6-aa+b.
2(b- a) As P(X<45) =0.31
i.e. x= 45 should be on left hand side of mean u and
Also t=2
we must.have
b -a' (b-a)X(b' +ab +a')
3(b-a) 3(b- a) ..)
1ean 6'+ ab +a'
Also P(X >64)=0.08
= i.e: X=64 should be right hand side of the
2018]
and we must have mean
dom :. Variance =h-K
ular .ii)
b'+ ab+ a
by:
3
Now P(X< 45)=0.31 =» P(Z<-Z})=0.31
»P-Z <Z<0) =0.19
4b +4ab +4a-3a-6ab-3b »P(0<Z<Z) =0.19
12 »Z =0.50
 B.Tech (M Sem.) C.S. Solved Papers
AEM.16
: Angle between two regression
lines
Also P(X> 64) =0.08 = P(Z>Z,)= 0.08
tan9 =-M
’ P(0<Z<Z)=0.42
1+ m,M,
+1. It means
The correlation coefficient (r) Iie between -l to
From equation (i):. and (ii) 0.50=-E and the value ofr denoted by modulus of r.
Angle between regression lines
1.41-04-#
dividing above, we get =49.9 ox +oy
14
and 14lg=64-u ’g==10
14 tan
Hence X-N(50, 10) line in
Significance : When one variable of regressionpositive.
is
creases as the other increasès the correlation
increases it is
When one variable decreases as the other
PART-C .negative.
r=0shows the absense of corrélation.
a
the two line.of Q18Applying the theory of least square method, fit
Q.17 If be the acute angle betveen second degree parabola to the following data :
regression of variables x andy, show that
3 4

tan = where i, ox, oy have. their 10|22 38

ar+oy
r=0 .RTU2023}
usual meaning. Explain the signiftcance where
{RTU2024]
and r +1,
Ans. We know that second degree of parabola equation is
Ans. Regression line of y on x given by y=a+bx +cx
and the normal equations are
-7=s-) }y=an +bEx+cx? .)
y-p= ..) Exy-ax+ bEx? +cx?
Compare with y=m+c Exy=aZx?+bEx' +ex
n=5
roy
Regression line of xon y
I
ro (y-) 10 16 20 40

3 22 .9 27 81 66 198
(*-)=(y-)
rãX
38 16 64 256 1S2 608

..(1) Ex=10 Iy= 76 30 L0o354 Zy=243Exy=&s1

y= rox -+ù
r¡x
Putting the value in equation (i), (ii) and(ii)
Compare with y=mx +c 76 = ax5+ bx10 +Cx30 i(iv)
243 ax 10+b x 30 + c x 100
m, =
 Advanced Engineering Mathematics AEM.17
Solved Papers
851 =ax 30 + b x 100 +c x 354 ..(vi) It is clear that x =75 andx= 80, tboth are on the
On solving the equation (iv), () and (vi) we get the right side of x=
values of a, b, c as,
Therefore let for x=75,
a6,31
itÍ +1. It means 7

Putting the values ofa, b and c in ... (1)

y=a+bx + cx?
10, 17. and for x=80,
y

... (2)
Q.19 Calculate the coefficient of correlation from the
folowing data: Hence
ession line in X: 2 3 4 5 6 7 8 9
P(x < 75) =0.58
on is positive. Y:98 1012 11 3 14 16 15
increases it is ’ P(z<z)=0.58
Also obtain the equations of line of regression andobtain »0.5 + P(0 <z<z) = 0.58
an estimate of Ywhich should correspond on the
Tverage to X= 6.2. |RTU 2022] ’ P0<z<z) =0.08 Igiven f(0.20) =0.08]
nethod, fit a Z, =0.20
ata : Ans. Refer to Q.8
and P(x> 80) =0.04
As we requireyat x = 6.2 hence line of regression
ofx givs us the best estimate as » P(z> z) =0.04
y=.95 x+7.25
’ 0.5- P(0 <z<z,) =0.04
y=95 x 6.2 +7.25
RTU20231 y= 13.14 at r=6.2 » P(0<z< z,) =0.46 Lgiven f(1.75) =0.46]
’ Z=1.75
equation is Q.20 lfthe skulls are classified as A,Band Caccording Using equation (1) and (2) we get
as he length- breadth index is under 75, between 7S 0.20g =75-p
and 80, or over. 80.: Using.normal distribution find ..(3)
.approximately the mëan and standard deviation ofa and 1.75g=
..) series in which A are 58%, B are 38% and Care 4%, 80 ... (4}
being given that :
) Dividing them we get
=74.4
..(üü)
then f(0.20) 0.08 and f(1.75) = 046. RTU 2019] andsabtracting them we get a =3.5
ry
Ans, Let the random variablex denote the length and 0.2 Find mean and variance of Poisson distribution.
breadth index with mean u and standard deviation o.
JRTU 2019]
P(x<75) =0.58 and P(x>80) =0.04
40
) Ans. Refer to.Q.9.
198
608
Q.22) Define Poisson distribution and finditsmean and
Ex'y=&51 Variance. {RTU 2018]

.(v) x=75 x=80 Ans. Refer to 0:9.

..()
Fig
 0.4
eng

Writ

HIsTORICAL DEVELOPMENT 3 Ans.

Field
1.I

2. F
PREVIOUSaVEARSQUESTIONS 3. F
4. F
Ve
often the minimun or maximum ofa function, within agiven
set of constraints. Optimization plays a crucial role in
PART-A engineering, where it's used to design systems and processes 5, In
that achieve the desired outcome with the most efficient to
use of resources. the
Q.1 What is optimization technique? Give example. 6. In t
RTU2024] to th
Q.3 What is diferece between a sBack and surplus 7. In t
Ams. Optimization Techniques :Optimization techniquevariable? [RTU 2022] equi
is a mathematical approach used to find the best possible 8. In de
solution, or an optimal solution, for agivén problem. The carry
goal is to inaximize or minimizea particular objective Ans. Difference between a Slack and Surplus VariabBe : 9. In des
function while satisfytng any constraints. Optimization tower
techniques are widely used in various fields, including Aspect Slack Variable Surplus Variable 10. In des
economics, engineering, finance, machine learning, and Avariable added to A variable subtracted to
11. In solv
operations research. -cönvert a"s" (less convert a"2"(greater 12. In cont
Definition than or equal to) than or equai to)
Example of Optimization-Technique: inequality constrainit inequality constraint into of prod
Linear Programming: Suppose a factory produces two into an equation. lan equation.
products, P, and P., and wants tomaximize profit. Each Q3 Write ste
Represents unused. Represents èxcess over a
product has specific resöurce requirements and yields 'a resóurces in the minimum reqúirement in
certain profit. The factory has limited resources fo, Purpose system. the system.
production. Appliedto."s" Applied to "2" Ans. Histor
|Constraint.
Objéctive : Maximize the profit from products P, and ,. lconstraints (less than constraints (greater than research was
Let Type Jor equal to). or equal to). (1939-1943).
x= number of units of P, produced Matheinatical Added to the left side Subtracted from the left mathematici£r
y= number of units of P, produced Form
of the constraint side of the constraint how to best
Profit Fúnction: lequation. equation. requiring the ai
Maximize Z=5x+8y A zro value JA zero value indicates of various prob.
lindicates the khe constraint is binding, so effective their
Where 5 and &are the profits per unit of P, and P,, Value constraint is binding, meaning there is no -The methods de
respectively. |Interretation meaning no resources lexcess over the adopted 3y indu
lare left unused. ninimum requirement. operation researc
92 What is optimization? (RTU 2023, 19)
Indicates the amount Indicates the amount by
which a constraint
Significance by which aconstraint exCeeds the minimum Q.6An animalfo
Ans. Optimization : Optimization refers to the not fully utilized. mixture contai.
requirement.
mathematical process of finding the best possble solution,
 Adoaneeç bngincoing Natiieniatics {AEM.19)
Q.4 Write two applications of Ingredient A, costs 3.00 per kg and A, costs 8.00
engineering. optimization in per kg. Not more than 80-kg of A, can be used and at
(RTU2022) least 60 kg of A, must be used. Formulate the problem.
OR
Write four engineering applications of (RTU 2018)
optimization.
3 Ans.
Fields:Applications
of Optimization in
1. In finding the optimal
trajectories of
(RTU 2019)

Engineering
space
Ans. Let X, and x, be the number of units produce by the
company of type A, and A, respectively.
Objective function
Minz =3x, +8x,
"and missiles. vehicles Subject to conditions
2. For the vptimal design of the control X t X, =200
3. For the optimal design of the electrical systems. X S80
4. For the optimal design of networks. Xy 260
ithin a given air-crafts, aerospace
vehicles, rockets, with regard to their weight and and
ncial role in consumption of fuel, their speed, atmospheric effects o.7
of ternperature, pressure variations etc. Define the slack, surplus and artificial variables in
nd processes linear programming problem.
5. In the designing of frames and structures with (RTU2018]
ost efficient regard
to their strength and the quantity of material used in
their designing. Ans. Slack Variables : The positive variable which are
6. In the designing of newtools and machines with regard added to leît hand side of the constraints to convert them
to the new dimensions added in their working. into equation (equalities) are called slack variables.
nd surplus 7. In the designing of pumps, turbines, heat Ex: X +X, +x,s4
[RTU 2022 equipment etç. transter,X, X» x, >0
Can be converted to
8. In designing scooters, cars, heavy and light goods
sVariable :
carrying vans.
9. In designing of foundations of big buildings, bridges, then x, is called slack variable.:
Variable
towers, dams, chimneys etc. Surplus Variables : The po_itive variables which are
10. In desigring the optimal pipe line network. substracted from the left hand side of the constraints to
sbtracted to
" (greater 11.In solvingthe travelling salesman problem. convert them into equalities are called surplus variables.
to) 12. In controlling the idle time and waiting time in queues Ex. X +x, 2 200
nstraint into
of production lines to reduce the costs. Can be converted to
X+X-Xg =200
Xcess'over a Q.5 Write short note on history of optimization. then x, is called surplus variable.
[RTU 2018]
quirement in Artificial Variables : Artificial variable are added to those
constraint with equality (=) and greater than or equal to ()
Ans. History of Optimization: The word operational sign. An artificial variable is added to the constraint toget
war-l! ;an initial solutionto an LP problem.
yeater than research was came into existence during world scientists,
(1939-1943). At that time the various team of
rom the left mathematicians, economist wereconstitutedto deciamgas0.8Á carpenter has 90, 80 and 50 running feet
design and operate man-machine system
onstraint how to best respectively of teak plywood and rosewood. Product A
requiring the allocation of scare resources. The solutions. equures ky 1 and 1 running feet of teak,; plywood and
indicates of various problems, suggested by the teams, proved to be
at is binding, winning the war. rosewood respectively. Product Brequires I, 2 and I
so effective their adoption ultimately led to teams anciently|runnn3 Jeet o teak, plywood and rose wood
re is no The methods develop by those research andB would seli
the adopted 'y industry and society after and now known as especrvely fa would sell for Rs. 48
techniques. for Rs. 40per unit, how much of each should he make.
quirement. operation research network or optimization and sell in order tà obtain the maximum gross incote out
amount by formuilation to
straint company must produce 200 ke ofa S Stock of w0od ? Give a mathematical (RTU 2017]
minimum Q.6 An animal food this linear programming problem.
mixture containing ingredients A, and A, daily.
 AEM.20 B.Tech. (M Sem.) C.S. Solved Papers
Atvancd
Ans. Let:
A, B and
Product mathematic
Teak Max
14
Plywood 2 Subi
Rosewood 1 1 5
Selling price per unit Rs. 48 Rs. 40
Let x, and x, be the number of units of the
Aand Bto be made respectively. products
can be formuiated Then the above problem Q,11 Find
mathematically
Max Z- 48 x+40 x,
as : multivaria
For optimization Mit
Subjectto 2x, +x, s90 Sub
X +2x, s 80 Z=t+y'+?
Xtx, S 50
Xj, X, >0 Ans. We u

196 49
PART-B 25 25
245
-=9.8
25
Q9 Optimize Z=+j+2
Q.10 A company desires to devote the excess capacty
Subject to: 4x +y?+2%= 14 RTU 2024| of the.three machines lathe, shaping and milling to
make three products A, Band C. The available time.
Ans. Given, per month in these machines are tabulated below : OL
-=2(
Z=t+'+ Machine
Available time
Lathe
200
Shaping Mlling
110hours 180 hours
Let g(*y,z)=4x+y' +2:-14=0 per month hous ’2x,=
ByLagrange multiplier equation The time (in hours) taken to produce each unit of
the products A, B and C on the machinés
L(r,y,2,2) =t+++(4x+y +2z-14) ..0) in the table below:
is di_played
P.D. wrt. X ßL
Machine Lathe
Shaping Milling =20
=2x+4 =0>=-22 Product A 2
Product B 2
’2x=37
NU
OL Product C 3
=2y +12y =0»y(2+12)=0 Nil 3
The profit per unit of the products ; B
and C are GL
y=0,2=-l 20, * l5 and 12 respectively.
Formulate the
mathematical model to maximize the profit.
2z+21=0’z=-1 .RTU 2023, 19)

OL Ans.
=4x+y' +2z-14 =0 Machine -252.
Lathe Shaping Milling Profit
(per 332
44-2) +(0 +2(-a)-14=0 Product A unit)
2 4 20
-82-21-14 =0 Product B 2 2 Nil 15
=-7/5 Product C 3 Nil 3 12
Available time 200 10 180
X=-22
per month hours hours hours
Adancod Ciyicering Kzthemctics AEM.21
Let x, X, and x, be the number ofunits of product
A, B andC to be manufactured respectively, then the -5x22
33 140-66 14 37
mathematical model is given by
2 66 66. 33
Max z=20x,+ 15x,+ 12x3
Subject to Sx,+ 2x,t 3x, s200 3x
-28 28
2x,+2x, +Ox,<110 +2 +2
33 -28+ 22 -3
4x,+0x, +3x,s 180 2 2 22 11

Q,11 Find the optimum of the following constrained So Min Z

multivariable problem :
Minimum Z=x+, + I t (ij-1)' =0.1799 + 4.4995 + 1.6198
Subjectto x, + 5t, -3x, = 6 =5.9394
[RTU 2019)
Q.12 Afirm manufacturing two types of electric tems.
Ans. We use Lagrange's multiplier method Aand B can nake a profo of Rs. 20 per unit of Aand
Rs. 30 per unt of B. Each unit ofA requires 3 motors
=-x+(x, +1)' +{%-) +(x, +Sx, -3xj -6) and 4transformers and each unit of Brequires 2and
4 respectively. The supply of these per month is 210
ôL
2x,+=0 and 300 respectively. Type B requires a stabilizer with
.supply of 65 units per month. Formulate the LPP for
maximum profit and _olve it graphically. [RTU 2017

Ans. Let the quantities of two types of electric items A

OL
and Bproduced are and respectively.
-=2(x,+ 1) + 5)=0 According to the question,
Max z=20x +30x,
’2x*-5),-2 s.t. 3x +2x, s210
4x + 4x, s300.
-S-2. And x, X, 20
2.
.Convertinggiven inequalitiesinto equation, we get
OL 3x +2x,-210 ...)
=2(X3-)-320
X; 0 70
’2x*3+2 X2 105 0
4x, + 4x,=300 ..)
Xg= 32.+2 C’(0,75)
2
A-’(70, 0)
oL
=X+5x,-3x,-6= 0 For point B,
2x equation (i)-equation (ii) ’
6x +4x, =420

’-252- 10- 92-6- 12 = 0

(32+2|-6=0 4x +4x, =300

33A -28 =0 2x1. =80

80
Or -=40
33
Froni,
-28 3x 40 +2x,= 210
XË=
66 2x, =210120 = 90
AEM.22 B.Tech. (m Sem.) C.S. Solved Pupers
(i) Design of water resources system for maximum
90
==45 benefit.
2 (iv) Design of pumps, turbines anú heat transfer
.:. B’(40, 45) equiprment for maximum efficiency.
Z, =20 x 70 +0= 1400
3. Minimization:
Z =20 x 40 + 30 x 45
=800+ 1350 =2150 () Planning of maintenance and replacement of
equipment to reduce operating cost.
Z, =20 x0+30 x 75= 2250
() Controlling the vesting and ideal times artd query
Hence Zm,= 2250 and. maximum point is B. in production line to reduce the cost.
(D Design of aircraft and aerospace structure fot
PART-C minimum weight.
(iv) Minimum weight design of structure for enter
quake wind and other types of random loading.
(v) Selection of machiningconditions in metal cutting
Q.13 What are the engineering application of process for minimum production cost.
optimization also give various classification of
optimization problems. (RTU 2024) (vi) Design of material handling equipment such,as
conveys trucks andcranes for minimum cost.

Ans. EngineeringApplications of Optimization: Refer 4. Management :

to Q.4. ) Deciding the locations site and industry.
Other Applications of Optimization Technique in () Planning the best strategy in acompetitive market
Engineering : Some of the typical applications from and devising ways of inereasing the earnings of
different engineering displace are given1below:
b company.
1. Genera : (I) Finding the shortest route which a salesman
) Optimal production planning, controlling' a should adopt 'so as minimize the cost and time
scheduling, spent in visiting anumber of cities.
) Inventory control. (iv) Inventory control to know when to purchase.
() Optimum design of control systems: How much to purchase to meet the future
(iv) Shortestroute taken of salesperson visiting various demands which will arise.
cities during one tour. (v) In the probability and statically fields.
() Optimum design of decimal machinery such as. Classfication of Optimization Problems:
motors, generator and transformer. 1. Based on the Nature of Design Vectors : Based on
(vi) Optimum design of electrical network. the nature of designvectòrs, optimization problems can be
(wi) Optimum design of chemical processing classified intotwocategories:
equipment and plants. () Static Optimization Problem : In this category,
(viüi) Design of optimum pipe risk network for. the problems are to evaluate a set of design
processes industries. paramters which makes the prescribed objective.
() Optimum design of linkages CAMS, gears, function of these parameters minimum or
machine tools and other mechanical components. maximum subjectsto the constraints.
(x) Design of civil engineering structure such as (i) Dynamic Optimization Problem: In this case,
frames, fundarmental bridges, towers, chimneys the objects isto obtain aset of design parameters
and dams for minimum cost. :
which are all continuous functions of some other
(xi) Finding the optimal targets of spare vehicles. parameters that minimize or maximize the object
(x) Optimal plastics design of structures. function subject to the constraints.
2. Maximization : 2. Based on the Nature of Expressions Involved : In
) Allocation of resourcesat services among several this section, according to the nature ofexpresions for the
activities to maximize the benefit. objective function and the constraints, optimization problms
(ü) Planning the best strategy to obtain maximum can be classified as linear, non-linear, geometric and
profit in the process of acompetitor. quadratic programming problems.
 AEM.23
.Papers quadratic
Advanced Engineering Matematics negativity conditions is called a ofgeneral
aximum () Linear Programming Problem : The programming problem. The statement
mathematical programming problem, in which the quadratic programming problem is
transfer objective function and allconstraints are linear
Minimize
and functions of design variables, is called as linear
programming problem. The general form of linear f()-2£,x;+ +c
ment of programming problem is Minimize or Maximize. j=l js1.
subject to
id query f(x)-ex i=l
j=1,2...,.m
subject to
ture for i=1,2,..n
j=1,2..,m X;20 constants.
Where C, Qui, Ci, a; and b; are all
Values of the Design
or entet
3. Based on the Permissible design
Variables: According to the values perrnitted for the
i=,2,...,n
bading.
as integer
Icutting Where c, a,, and b, are all constants. variables, optimization problems can be classified
() Non-linear Programming Problem:Ifany of and real-valued programming problems. some or
such as the functions f(X), g(X)S0,j=1,2,..,mandh (i) Integer Programming Problem :If...,x, of
cOSt. (X) = 0, j= 1,2,...,.p are non-linear, then the all of the design variables X,,X),
programming is called non-lincar programming optimization problem are restricted to take only
problems. integer values, the problem is called an integer
(ii) Geometric Programming Problem: A programming problem.
market function h(X) is called posynomial ifthe function (ü) Real-valued Programming Problem : Ifall the
ngs of is expressible as thè sum of power terms of the design variables are permitted to take any real
form value, the optimization problem is called real
esrman valued programming problem.
d time 4. Based on the Number of Objective Functions :
Where c; and a; are constants. with C; >0 and . Depending on the number of objective functions to be
x; > 0. Thus, amathermatical programming minimized, óptimization prob<ems can be classified as singie
chase.
and multi-objective programming problems.
problem, in which the objective function and al!
Future
constraints are expressed as posynomial in X, is () Single-objective Programming Problem:
known as geometricprogramming problem. The Single-objective programming problem in which
general form of geometric programming problern there is only a single objectiv.
is Minimize. (ü) Multi-objective Programming Problem: A
ed on multi objective progr£mming problemi can be
can be statd as follows:
s(x) = Find Xwhich minimizes f(X), f(X) ... X)
gory, c; >0, x>0 Subject to
esign g(X)s0,j= 1,2...,.m
ective subject to Where f h..s denote the objective functions to
m or be minimizedsimultaneousty.

case, Q.14 Write a short note on the classification of

1eters a_ >0, x, >0, k=1,2,...m. optimization problems based on various parameters.
other Where N, and N, denote the number of [RTU 2023, 22]
posynomial terrns in the objective function and OR
bject
k constraint function, respectively. How optimization problèm are classified based on the
i: In (iv) Quadratic Programming Problem : A non nature of expressions? .RTU2019]
or the linear programming problem with a quadratic
bléms obiective function and linear constraints with non Ans. Refer to Q13.
and
 Advanced Engineering Mathematics

solution of tl
0.2 Find all the basic
2x +y -z=2
3x +2y+¿=3

2*+y-z=2
CLASSICAL OrTIMZATION UsING Ans.

DIFFERENTIAL CALCULUS 4 3x+2y+z=3

PREVIOUs YEARS. QUESTIONS

which is equal to r

Now r = = 6x, +i8 Rank of matrix-2

PART-A . Equation haye basic
solutionn
-=0 Putting z=0 in equations
2x+y=2
Q.1 Find the maximaand minima of
t= =-6x, +36 3x+2y =3
++;+1a +14 RTU 2024, 16)
) AtA (0, 0), rt-s
Ans. Given function = (6x+ 18) (6x,+ 36)-0
-(0+ 18) (0 +36)
f=x+x+9x +18x, +144 =648> 0
at maxima or minima both with r=18>0
So minima at A(0, 0) is fmin144
f () At B(0, -12), rt -s 12 -][2
= (6x, + 18) (6x, +36) -0
=(0+ 18) (6 x-12+36) 2.
af =18(-36)
=0 =-648<0
SO
Ôx, =3x;+18%, Bis a saddle point.
SO x{(3x, +18) =0 () At C(6,0), rt- Basic soultion is x=1,
-18
=(6x, + 18) (6x, +36)-0
x=0 or x = 3
=(6 x-6+ 18) (0+36)
=-18%36
--648 <O
Again
h-13-4
We can get a basic solution at
and =3x, +36x, =0 Cis again a saddle point. yz*2
(iv) At D(6, -12), rt-s 2y +z=3
SO. x(3x, +36) =0 = (6x, + 18) (6x, +36) -0
= (6 x-6+ 18) (6 x -12 +36)
-36 -12 =(-18) (-36)
X, =0or X, = 3 =648> 0
with n=6x-6+ 18
So the points where extremes may be occur.
-18<0 (negative)
) A(0, 0) So there is a maxima at D with
() B(0,-12) fnax(-6)+(-12) +9(-6} +18(-12y +144
(Gi) C(-6,0) 144
fany -216-1728 + 324 +2592 + 12 1
(iv) D(-6,-12) =|16
2+2\-2 1
 Aanced Engineerlng Mathematics {AEM.25

0.2 Find all the basic solution of the system:

2x+y -z=2 44+3]
{RTU 2024]

Ans. 2*+y-z=2

4 3x+2y+z=3
: Basic solution

2 -1|
7 -1

Again 3 2+340
which is equal to number of equation. We can get abasic solution at y 0.
2x-z=2
Rank of matrix =2 3x+z=3
: Equation haye basic solution n,=3, =3
Putting z=0 in equations
2x+y=2
3z+2y =3

4-3|-3
.Basic solution is x=1, y=0,z=0.

Basic soultion is z= 1,y= 0, z=0. Q.3)Under whatcircumstances can the condition

j=0notbe used tofud the minimum ofthefunction
f) ? (RTU2023]
Again 2 1 =1+3=4=0
We can get a basic solution at x = 0 Ans. We know that f'(x)=0 is anecessary condition for
y-z=2 calculation of maximum and minimum value of function.
2y +z=3
Also, for minimum value, we calculate f°(x) and
check the value of f"(x) at point (xy). If f"(x) >0 then
function f(x) is mininum.

QWhat is Lagrangian function? (RTU2023,22]

44 Aus. Lagrangian Function : In optimization technique,

the Lagrangian function is &strategy to finding maximum
and minimum value ofa function which subjected to equality