LAB AcTIVITY –16

(Maxima and Minima)

TOPICa Applications of Derivatives

Meterial Required
(Objective cellotape/
Chart papers, scissors,

To construct an open box of maximum volume from a given adhesive, calculator.

rectangular sheet by cutting equal squares from each corner.

Procedure

1. Take a rectangular chart paper size 20 cm x 10 cm and namne it as

of
ABCD.
2. Cut four equal squares each of side x cm from each corner A, B, C
andD. (SeeeFig. 16.1)
Repeat the process by taking the same size of chart papers and different
3.
values ofx.

4. Make an open box by folding its flaps using cellotape/adhesive.

DX

10 cm

20cm
Fig. 16.1

Demonstration

1. When x=1, volume of the box 18 x 8 x 1= 144 cm

357
2. When x= 1.5, volume of the box 17 x 7x 1.5 = 119x3
2
-= 178.5 cm

3. Whenx= 1.8, volume of thebox = 16.4 x 6.4 x = 188.9 cm.

1.8

4. When x= 2, volume = x 6 x 2= cm.

of the box 16 192

5. When x= 2.1, volume the box = x 5.8 of =192,4 cm 15.8 x 2.l

x 5.6 x 2.2 =192.2 cm.

6. When x= 2.2, volume the box = of 15.6

7. Whenx=2.5, volume the box =15 5 2.5 = 187.5 cm.

of x x

8. Whenx=3, volume the box = 14 4 3 =168 cm.

of x x

Clearly, volume the box is maximum when x =2.1.

of
Lab Manual in Mathematics-XIl /40
 Observation
1. V,= Volume of the open CNOTE
box (whenx=1.6)-
2.
V2=Volume the open box Let V denote the volume ofthe box.
of
(whenx= 19) -
3, V=Volume Now V= (20- 2r) (10-2r)x
4.
of the open box (whenx=2.1) = or V= 200x- 60x?+4r
V=Volume ofthe open box(whenx=2.2)=
5, V4=Volume of theopen box (whenx- 2,4) = dr
= 200 - 120x + 12x².
6. Vs=olume of the open box (whenx= 3.2) = For maxima or minima, we have,
7. Volume V is
than volumeV.
8. Volume V,is =0, i.e., 3r-30x +50 =0
than volumeV.
9. Volume V3 is
than volumeV. i.e.,

10. Volume V4 is
than volume V. 30tV900–6001.9 or 2.1
11. Volume V5 is 6
than volumeV.
Thus, volume of the open box is Reject x =7.9.
maximum when x=2.1.
d'v
Application =-120 +24x
d
This activity is useful in explaining the
concepts of maxima/minimaof
functions. It is also useful in making packages of
maximum volume with
When x =2.1, is negative.
minimum cost.

D Some useful Results For Solution of Practical Problems Hence, V is maximum x =2.1.

at
1. Area of Square (side x) =
2. Perimeterof Square (side x) =4x
3. Area of Rectangle (sides x and y) =y
4. Perimeterof Rectangle (sides x and y) =2(x +y)
1

5. Area ofTrapezium =,(sum ofparallelsides) x (Distance between them)

6. Area of Circle(Radiusr)=

7. Circumference of Circle (Radius r) =2r

8. Volume of Sphere(Radiusr)

9. SurfaceArea of Sphere (Radiusr) =4m

10. Volume of Right Circular Cylinder(Baseradius r and height h) =nrh
11. SurfaceArea of Right Circular Cylinder =2rh+2n where base radius =r and height =h
12. Curved Surface Area of Right Circular Cylinder =2rrh

13. Right CircularCone of Base Radius r, Height h and Slant Height !

Volume =, r'h
Curved surface Area =rl, Total Surface Area = +rl

Lab Manual in Mathematics-XIl /41

14. Volume of Cuboid (Length of edges x, y, z) =xy z, oTE
Surface Area =2(y +yz+zx) If k is a positive constant, then:

=,
15. Volume of Cube (Length of edgex) Surface Area =6r function

be
), ),
maximum
of the form k fr),
C)",
or
log
k -

fr) will
minimum
16.Area of EquilateralTriangle =(side)?
according as f() is maximum or
minimum provided thatfr) >0.
17. Area of Triangle =ab
2 sin C or bc
2 sin A or ca sin B

18. Area of Triangle = s(s-a) (s-b)(s-c)(Hero Formula)