APPLICATION OF

DIFFERENTIATION
4.2 Extremum Problem
 OBJECTIVES :

At the end of the lesson students will be able to

a. Find the stationary points.

b. Identify minimum and maximum points by using

first and second Derivative Tests.

c. Solve optimization problems .

Introduction

How do they design roller coasters? How do they figure

out the shapes, how do they know how steep to make
the angles, and how do they know how fast it will go?
The answer is, of course, mathematics!
 Critical
Point

Stationary point if
f ' ( x)  0 f ' ( x) does not exist

Maximum Minimum
point point
To find the stationary point

Definition

Stationary A number c in the domain of

Value f(x) such that f’(x) = 0
Stationary x = c , y = f(c)  (c,f(c))
Point **when f’(x) = 0

5
 Stationary point

y f(x)
(c,f(c))
f’(c)=0 f’(d)=0

(d,f(d))
x
0 c d

• the tangent to the curve is horizontal

• the gradient of the curve is zero
6
  maximum point
Two types of stationary point
 minimum point

as shown in the next diagram

dy a = stationary
f (a)  0 or  0 when x  a
dx value
Maximum point Minimum point

a a
 Example 1
Find the stationary points for the function
a) f(x) =12x-3x 2

b) f(x) = 2x + x -20x +5
3 2

stationary points :
f’ (x) = 0

8
8
 Solution
a) f(x) =12x-3x 2

(2,12)
when f'(x) = 0
stationary
12-6x = 0
point
6x =12
x =2 stationary
value
when x = 2
f(2) =12(2)-3(2) 2

=12
9
b) f(x) = 2x + x -20x +5
3 2

when f'(x) = 0:
6x +2x-20 = 0
2

3x + x-10 = 0
2

5
(3x-5)(x +2) = 0 x = or x =-2
3
Stationary values

10
 Example 2
Given a function y  5x  5x  1
2

Find the stationary points of f (x).

Solution 1
When x
y  5x2  5x  1 2
2
1 1 5 5
dy y  5    5  1   1
when 0 2  2 4 2
dx 1

10 x  5  0 4
10 x  5 1 1
  ,   is a stationary point.
1 2 4
x
2
 Function Increases and Decreases

Definition & Theorem

a) f is increasing in interval I if

i) f(x1) < f(x2)

f(x2)
ii) f ’(x) > 0
f(x1)

x1 x2
b) f is decreasing in interval I if

i) f(x1) > f(x2)

ii) f’(x) < 0

f(x1)

f(x2)
x1 x2
c) f is a constant in interval I if

i) f(x1) = f(x2)

ii) f’(x) = 0

f(x1) = f(x2)

x1 x2

Note : interval I = interval (a, b)

First Derivative Test
 b) The First Derivative Test

Suppose that c is a stationary number of a function f that is

continuous on (a, b).

a. If f ‘ (x) > 0 for (a, c) and f ‘(x) < 0 for (c, b), then f has a
relative maximum at c

b. If f ‘(x) < 0 for (a, c) and f ‘(x) > 0 for (c, b), then f has a
relative minimum at c

c. If f ‘(x) does not change sign at c, then f has

no relative extremum at c
It easy to remember the First Derivative Test by visualizing
diagrams such as those in Figure 3.22
f(x) f(x)

f ‘(x) > 0 f ‘(x) < 0 f ‘(x) < 0

f ‘(x) > 0

x c
c
b. relative minimum
a. relative maximum
f(x)
f(x) f ‘(x) < 0
f ‘(x) > 0

f ‘(x) < 0
f ‘(x) > 0
x
x
c
c. No extremum d. No extremum
Interval Determination

S1 : Determine the domain f(x)

S2 : Determine the interval to test
S3 : Test the f’(x) sign that determine in S2
S4 : Conclusion
 Given a function y  5x2  5x  1
Example 1
Find all the relative maximum and
relative minimum points of f (x).
Solution
1
y  5x  5x  1
2
When x 
2
dy
when 0 1
2
1 5 5
dx y  5   5  1   1
2  2 4 2
10 x  5  0
1
10 x  5 
4
1
x 1 1
  ,   is a stationary point.
2 2 4
 Using the first derivative test
1 1
 ,  is a stationary point.
2 4
 1 1 
Interval  ,   ,
 2 2 
Test value, x 0 1
Sign of
dy negative positive
 10 x  5
dx
Curve of y
1 1
  ,   is a minimum point
2 4
Example 2
Find the maximum or minimum values of the
function y = 4x 3 + 15x 2 – 18x + 7.

Solution
y = 4x 3 + 15x, 2 – 18x + 7
dy
 12x 2  30x  18
dx
dy
To find the stationary numbers we set 0
dx
12x 2  30x  18  0
6(2x – 1)(x + 3) = 0
1
So the stationary numbers are or -3.
2
To determine the maximum or minimum,
 ,3  1 1 
Interval   3,   ,
 2 2 
Value of x 4 0 1
Sign of f ’(x) + - +
Conclusion increasing decreasing increasing

dy
 12x 2  30x  18
dx
x  4 dy
 12 42  30 4  18
dx
 192  120  18
 54
You can see from the chart that dy changes from positive
to negative at x = -3. dx
So, by the First Derivative Test, f has a maximum at x = -3
and the local maximum value is

y = 4x 3 + 15x 2 – 18x + 7 when x = -3, y = 88

Since f ’(x) changes from negative to positive at x = 1 / 2, so f

has a minimum at x = 1 / 2 and the local minimum value is

y = 4x 3 + 15x 2 – 18x + 7 when x = 1 / 2, then y = 9 / 4

 Example 3 Given a function f ( x )  x 3
 3x  2.
Find all the relative maximum and relative
minimum points of f (x).
Solution
f ( x)  x  3x  2
3
f (1)  1  3(1)  2
3

When f ( x)  0 0
3x  3  0
2
f (1)  (1)  3(1)  2
3

x 1  0
2
4
 x  1 x  1 , 0 the stationary points
x 1 or x  1  1, 4  , 1, 0 
 the stationary points
Using the first derivative test  1, 4  , 1, 0 
Interval   x  1 1  x  1 1 x  
Test value, x 2 0 2
Sign of
positive negative positive
f ( x)  3x  3
2

Curve of f(x)
Since f’(x) changes from
relative maximum point  1, 4  positive to negative as we
move across x = -1

Since f’(x) changes from

relative minimum point 1, 0  negative to positive as we
move across x = 1
Example 4
Determine the interval of f(x) = x4  2x2 which the
function are increasing or decreasing.

Solution :
S1 : Domain, Df = (, )
S2 : Determine the interval by critical value
f’(x) = 4x3 – 4x
When f’(x) = 0, then 4x3 – 4x = 0
4x(x2 – 1) = 0
Thus, the critical values are
x = 0, x = 1 or x = -1
S3 : Test the f ’(x) sign

Interval (, 1) (1, 0) (0, 1) (1, )

Value of x 2 0.5 0.5 2

Sign of
f ’(x) negative positive negative positive

Conclusion decreasing increasing decreasing increasing

S4 : Conclusion
(, 1) & (0, 1) are decreasing interval and
(1, 0) & (1, ) are increasing interval
 Exercises

1.The curve y = x 2 + ax + b has a stationary point

.
at (1, 3). Find a and b.
2. For what values of a and b will the function
f(x) = x 3 + ax 2 + bx + 2 have a relative maximum
when x = -3 and relative minimum when x = -1.

3. Find a cubic function f(x) = ax 3 + bx 2 + cx + d

that has a relative maximum value of 3 at -2 and a
relative minimum value of 0 at 1.
4. Determine the interval which the function are
increasing or decreasing for the following function.
Then, sketch the graph.
Ans i) :
3 2 decreasing at (∞, 1) & (0, 1)
i) f(x) = x3  x +1
2 Increasing at (1, 0) & (1, ∞)

 x 2  1 , x  0
 Ans ii) :
ii) f ( x )  1 , 0  x 1 Increasing at (∞, 0) &
 2 (2, ∞)
x  4 , x  2
 b) To find the relative exremum using
the second derivative test
Second derivative can determine the maximum
and minimum values of a function

If f ' (c)  0 and f '' (c)  0, then f has local minimum at c

If f ' (c)  0 and f '' (c)  0, then f has local maximum at c

If f ' (c )  0 and f '' (c )  0, then the test fail, it is inconclusive.

Thus the first first derivetive test is used.
 Example 1
By using the second derivative test, find the relative

extremum of f ( x)  x 3  3 x 2  24 x  32
Solution
f ' ( x )  3 x 2  6 x  24
 3( x  2)( x - 4)

when f ' ( x )  0, then x  -2 or x  4

so the critical numbers are - 2 and 4

To determine the relative extremum,

the second derivative is used

"
f ( x)  6 x  6
 At x  -2, f " ( x )  18  0 and
at x  -4, f " (x)  18  0

When x  -2, then f (-2)  60

when x  4 then f (4)  -48

So by the second derivative test, (-2,60) is

a relative maximum of f and (4,-48) is
a relative minimum of f
 Example 2
Find the stationary points of the curve
y = 4x3 - 5x2 + 2x + 3 and determine whether each
turning point is a maximum point or a minimum point.
Solution
Stationary
y  4 x  5x  2 x  3
3 2
point : dy
0
dy dx
0
dx
12 x  10 x  2  0
2

6 x  5x  1  0
2
1 1
(3x  1)(2 x  1)  0 x  or x 
3 2
 1 dy
When x   12 x 2
 10 x  2
2 dx
3 2 2
1 1 1 d y
y  4   5   2   3  24 x  10
2 2  2
2
dx
1 d2y
13  1 13 
When x  , 2
 2  0
  ,  2 dx
4 2 4  d2y  1 13  Min
 0,   ,  point
 
2
1 dx 2 4
When x 
3 2
1
3
1
2
1 1 d y
y  4    5    2    3 When x  , 2  2  0
3  3  3 3 dx
 1 88  Max
2


88  1 88  d y
 0,   ,  point
 , 
 3 27 
2
27  3 27  dx
Example 3 Given that y = 12 x ( 2 – x ). Calculate the value
of x when y is a maximum.
Hence, find the maximum value of y
Solution
y = 12 x ( 2 – x ) when x = 1
y = 24x – 12x2 y = 24(1) - 12(1)2 = 12
dy
 24  24 x d2y
dx 2
 24  0
At critical point, dx
dy d2y Max
=0  0,  (1,12) point
dx dx 2
24 – 24x = 0
So the maximum value of y,
24x = 24
x=1 y = 12
Example 4
By using the second derivative test, find the relative

extremum of f ( x)  x 4  4 x3
Solution
f ' ( x )  4 x 3  12 x 2  4 x 2 ( x  3)
When f ' ( x )  0, then x  0 or x  3

" 2
f ( x)  12 x  24 x

At x  0, f " ( x)  0
The test fails,it is inconclusive.
Thus the first derivative is used.
Interval ( 0,3 )
(,0)

Test value of x -1 1

Sign of f ' ( x) - -

Conclusion
decreasing decreasing

Hence, the first derivative test shows f does not

have a relative extremum at x=0
 At x  3, f " ( x )  36  0

When x=3, f(3)=-27.

So by using the second derivative test,

(3,-27) is a relative minimum of f
Extra Examples
 Example 1 Find the relative extremum for the
function f(x) = 3x2 + 2

Solution 1) Find the stationary point (max / min point)

Using the first  0, 2  is a stationary point.

derivative test
Interval  , 0   0,  
f(x) = 3x2 + 2
when f ( x)  0 Test value, x -1 1
6x  0 Sign of
-ve +ve
x0 f '( x)  6 x

f (0)  3(0) 2  2 Curve of y

f (0)  2   0, 2    0, 2  is a minimum point
or 1) Find the stationary point (max / min point)
Using the second derivative test
f(x) = 3x2 + 2

when f ( x)  0 f ( x)  6  0
6x  0 f ( x)  0
x0
  0, 2  Minimum point
f (0)  3(0)  2
2

f (0)  2   0, 2 
2) The positions of the roots (if any) of f(x) = 0
x – intercepts Domain –2 < x < 3
when f ( x)  0 f (2)  3(2)  2  14
2

3x  2  0
2 f (3)  3(3)  2  29
2

2 f(x)
x 
2

3 29
2
x 
3 14
(undefined) 2
f(x) does not cut at x-axis x
–2 0 3
3) y – intercepts The range of f (x) is
f (0)  3(0)  2  2
2 2  f ( x)  29 or R f  [2,29]
Example 2
Find the maximum or minimum values of the
function f ( x)  3x 3  3x 2  3x  2
Solution:

Process to find the stationary values

(max or min point)

f ( x)  9 x  6 x  7
' 2

(3x  1)(3x  3)  0
1
x  @ x  1
2
 If by using first derivatives test
intervals (,1) 1 1
(1, ) ( , )
3 3

test value -2 0 1
sign of f’(x) +ve -ve +ve
conclusion Increasing decreasing increasing

Relative maximum:
f (1)  3(1)3  3(1) 2  3(1)  2  5, (1,5)
Relative minimum:
1 1 1 1 4
f ( )  3( )3  3( ) 2  3( )  2  1
3 3 3 3 9
If by using second derivatives test

f ' ' ( x)  18 x  6
f ' ' (1)  18(1)  6  12 ,  0 max
f ' ' (1 / 2)  8(1 / 2)  6  10 ,  0 min

f (1)  3(1)3  3(1) 2  3(1)  2  5

1 1 1 1 4
f ( )  3( )3  3( ) 2  3( )  2  1
3 3 3 3 9
 5

2
3
39

4
1
9 1

-2 -1  1 1
3 3
 Exercise :
1. For the function y = f (x) below, find all the
relative maximum points and minimum points
by applying the second derivative test.
(a) f ( x)  x 2  2 x  24
(b) f ( x)   x3  9 x 2
(c) f ( x)  x3  3x

2. Sketch the graph of f ( x)  27  3 x 2

for 4  x  6 and state the range of f (x).

3. Sketch the graph of f ( x )  3 x 2

 7x  4
and state the axis of symmetry.
Answers :
1. (a) ( 1, –25 ) is a minimum point

(b) ( 0, 0 ) is a minimum point ;

( 6, 108 ) is a maximum point

(c) ( 1, –2 ) is a minimum point ;

( –1, 2 ) is a maximum point
 y
2.

27

-4 -3 0 3 6 x
-21

-81

The range of f(x) is 81  f ( x)  27

 c) Solve Optimization Problems

OBJECTIVE

Solve Optimization Problems

 Optimum Value Problems
A function y = f(x)

 is maximum at x = c when
f’(c) = 0 and f’’(c) < 0,
the maximum value is f(c);

 is minimum at x = c when
f’(c) = 0 and f’’(c) > 0,
the minimum value is f(c).
 Formulae
Area Volume
1.Circle A   r 2
1.Cuboid/cube
2.Sphere A  4 r 2 V  length  width  height
3.Hemisphere A  2 r 2
2.Sphere V   r
4 3
4.Cylinder A  2 r  2 rh
2 3
5.Cone A   r   rs
2 3.Cylinder   2
V r h
1 2 1 2
6.Sector A  r  4.Cone V  r h
2 3
7.Square A  length  width
 Steps For Solving Optimization Problem

1. Identify the variable to be maximized or

minimized, say variable y.
2. Express y into a single variable function,
say y = f(x).
3. Solve f’(x) = 0.
4. Verify the type of solutions of f’(x) = 0 by
using the first or second derivative test.
5. Get the optimum value y = f(x) if it is required.

57
 Example 1

An open-top box is to be made by cutting small

congruent squares from the corners of a 12 cm
by 12 cm sheet of tin and bending up the sides.
How large should the squares cut from the
corners be to make the volume of box
maximum?
x
x
12cm
 x

12-2x

12-2x

 V  x(12  2 x)(12  2 x)
Solution :
Volume of the box, V  x (12  2 x)(12  2 x )
2 3
V  144 x  48 x  4 x
dV
 144  96 x  12 x 2

dV dx
Let 0
dx
12 x 2  96 x  144  0
(12) x  8 x  12  0
2

( x  6)( x  2)  0
x  6 or x  2
 d 2V
For 2
 96  24 x
dx
2
 96  246  48  0
dV
when x  6, 2
dx

 V is minimum when x = 6.
2
 96  242  48  0
dV
when x  2, 2
dx

 V is maximum when x = 2.
Therefore, the area of each squares to be cut
from the corners, to make the volume of box
maximum is (2 x 2) = 4 cm2.
 Example 2
You have been asked to design a 1 liter oil
can shaped like a right circular cylinder as
shown below :
2r

2000
(a) Show that A  2r 
2
r
(b) What dimensions will use the least material?
Solution :
(a) Volume of the can, V   r 2 h  1000
1000
h 
 r2
Surface area of the can,
A  2r  2rh
2

 1000 
 2r  2r  2 
2
 r 
2000
 A  2r 
2
shown
r
 (b) 2000
A  2r 
2
r
dA 2000
 4r  2
dr r
dA 2000
When  0, 4r  0
dr r2
4 r  2000
3

2
0
r
500
r 
3


500
r  3  5.42 unit

 2
d A 4000
2
 4 
dr r3
When r  5.42,
2
d A 4000
2
 4 
dr 5.423
 37.69  0
 A is minimum when r  5.42 units
1000
Therefore, h   10.84 units
 (5.42) 2
Example 3 If L = x 2 y and x + y = 6, find
the maximum value of L .
Solution

Given L = x 2 y --(1) dL
0
and x + y = 6 dx
y = 6 – x --(2)
12x – 3x2 = 0
Substitute (2) into (1)
3x ( 4 – x ) = 0
L = x2 (6 - x)
L = 6x2 – x3 # x = 0 or x = 4
 2
d L
2
 12  6 x
dx
2
d L
When x = 0, 2
 12  6(0)  12  0 (minimum)
dx
2
d L
When x = 4, 2
 12  6(4)  12  0 (maximum)
dx

L is maximum Lmax  x y  (4) (2)  32

2 2

when x = 4,
The maximum value of L is 32.
y=6–x=2
Example 4

A cuboid has the following dimensions,

length = ( 10 – 3x ) cm, breadth = x cm
and height = x cm.
Calculate the maximum value for the
volume of the cuboid.
SOLUTION :
Volume of cuboid
V = ( x )( x )( 10 – 3x )
x cm V = 10x2 – 3x3 #

dV
10 - 3x cm 0
dx
20x – 9x2 = 0
x cm
x ( 20 – 9x ) = 0
20
x = 0 or x 
9
 2
d V
2
 20  18 x
dx
d 2V
When x = 0 , 2  20  18(0)  20  0 (minimum)
dx
 20 
2
20 d V
When x  , 2  20  18    20  0 (maximum)
9 dx  9
 20
When x  ,
9
2 3
 20   20 
Vmax  10 x  3x  10    3  
2 3
 9   9 
The maximum value
of the cuboid  16.461 cm 3
Example 5 S 60 cm R

x cm
M
40 cm

P N Q
3x cm
The above diagram shows a rectangle PQRS
such that SM = x cm and PN = 3x cm.
(a) Show that the area of the shaded region,
3 2
A cm2 , is given by A  2400  60 x  x
2
(b) Find the minimum value of A.
SOLUTION (a) Area of PQRS = ( 60 )( 40 )
= 2400 cm2
1
Area of PMN  (3x)(40  x)
2
3 2
 60 x  x
2
Area of the shaded region,
A = Area of PQRS – Area of PMN
 3 2
 2400   60 x  x 
 2 
3 2 (shown)
 2400  60 x  x
2
(b) The minimum value of A

3 2
A = 2400  60 x  x A is minimum when x = 20
2
dA
0 Amin imum
dx 3
 2400  60(20)   20 
2

-60 + 3x = 0 2
3x = 60  1800 cm 2
x = 20
2
d A
2
 3  0 (minimum)
dx
Example 6
Refer to the figure below, a cone has a circular
base with radius r and the height of the cone is h.
Given that the length of AB is 3cm, find the
value of r and h if the volume of the cone is to be
maximized. B

3cm h cm

r cm O
A
SOLUTION :
B OAB is a right-angled triangle
From Pythagoras theorem,
r h 3
2 2 2
3cm h
r  9  h
2 2

1
A
r O Vcone  hr 2
3
1
Substitute r 2
 9  h 2
into V  hr 2

3
1 1
V   h(9  h )  (9 h   h )
2 3

3 3 1
V  3 h   h 3

3
 1 dV
V  3 h   h 3
 3   h 2

3 dh
dV
0 2
d V
dh 2
 2h
dh
3   h  0
2
When h  3
 (3  h )  0
2
2
dV
 2( 3)  0
3 h  0
2
dh 2
(maximum)
h 3
The volume of the cone
Let h  3 is maximum when
because h is a positive value h 3
When h  3
r  9h
2 2

 9  ( 3) 2

r 6
2
Let r  6
r 6 because r is a positive

 The volume of the cone is maximum when

h  3 cm and r  6 cm
Example 7
The area of a rectangle is 36 cm2. Find
the minimum value of its perimeter.

Solution

Given area of a rectangle is 36 cm2

Let xy  36 x cm
36
y .........(1) y cm
x
The perimeter of rectangle
P  2 x  2 y .........(2)
(1) into (2) : P  2 x  72
x

dP
0 x  36
2

dx
72 x  6
2 2  0
x Let x  6
2 x  72
2
Since length cannot
2
 0 , x ≠ 0
be a negative value
x
2x2 – 72 = 0
2 x  72
2
dP 72 2
 2  2  2  72 x
dx x When x = 6
2 72
d P 144
 P  2x 
dx 2
x 3 x
72
when x = 6 , Pmin  2(6) 
2
d P 144 6
2
 3  24cm
dx 6
2
d P 2
2
 0 The minimum value of the
dx 3 perimeter of the rectangle
 x  6(minimum)
Example 8 10x cm

y cm y cm

13x cm 13x cm

The diagram shows a polygon that is formed

from a wire of length 96 cm. Show that the area
of the polygon, A cm2 is given by A = 480x – 120x2.
Find the maximum value of A.
Solution Perimeter of the polygon
10x + 2y + 2(13x) = 96
10x cm
36x + 2y = 96
2y = 96 - 36x
y cm y cm
5x cm y = 48 – 18x ---(1)
Area polygon
12x cm A= area rectangle + area triangle
1
13x cm 13x cm  10xy  (10 x)(12 x)
2 2
 10 xy  60 x - - - (2)
(1) into (2) : A  10 x(48  18 x)  60 x
2

 480 x  180 x  60 x
2 2

A  480 x  120 x (Shown)

2
Find the maximum value of A. The maximum value
dA of A :
0
dx When x = 2 ,
480  240 x  0
A = 480x – 120x2
240 x  480
Amax = 480(2) – 120(2)2
x=2
dA = 480 cm2
 480  240 x
dx
2
d A
2
 240  0
dx
 x  2 ( maximum )
 Example 9 x cm y cm

The above diagram shows two circles of radii x cm

and y cm respectively. If x + y = 10.
(a) Show that the total area of the two circles, A cm2
is given by A  2 x 2  20 x  100
(b) Find the value of x such that A has a minimum
value and calculate the minimum value of A.
Solution
(a) Show that the total area of the two circles, A
2
cm is given by A  2 x 2
 20 x  100
Radii of the two circles
Given x + y = 10
y = 10 – x ------ (1)
Total area of the two circles
A   x 2   y 2 ------ (2)
(1) into (2) A  x  (10  x )
2 2

 x  (100  20 x  x )
2 2

  x  100  20 x   x
2 2

 2 x  20 x  100 ( Shown )

2
(b) Find the value of x such that A has a minimum
value and calculate the minimum value of A.
A  2 x 2  20 x  100
dA
0
dx
4 x  20  0
4 ( x  5)  0
x = 5 → critical value
2
d A
 4   0  x  5 ( minimum )
dx 2
When x = 5, Amin  2 (5)  20 (5)  100
2

 50 cm 2
1. A wire of length 70 cm is formed into a sector
with a radius of r cm as shown below. r cm

(a) Express the value of θ in terms of r.
(b) Show that area of sector, A cm2 is given by
A = 35r –r 2 .
(c) Find the maximum area of the sector.

70
Answers (a)   2
r
2
(c) 306.25 cm
2. A right close cylinder has a volume of 500cm3.
Its base and radius is r cm.

(a) Show that the surface area of the cylinder, A cm2 is

given by
1000
A  2 r 2

(b) Hence, find the value of r at which A is a maximum.

Answers

(b) r = 4.3cm, A = 348.7cm

3. A closed right circular cylinder of base r cm
and height h cm has a volume of 54 cm
3

Show that S, the total surface are of the cylinder

is given by
108
S  2 r 2
r
Hence, find (i) the radius and height which make the
surface area is a minimum.
(ii) the minimum area.
4. A lump of modelling clay volume 72 cm3 is moulded
into the shape of a cuboid with edges of lengths x cm,
2x cm, and y cm. Calculate the minimum surface
area of this cuboid.
y cm

2x cm

x cm
Answers
A  108 cm2