Higher Engineering Mathematics, Sixth Edition

Some applications of differentiation 307

Now try the following exercise

Exercise 122 Further problems on turning

points

In Problems 1 to 11, find the turning points and

distinguish between them.

1. y=x^2 − 6 x [(3,−9) Minimum]


2. y= 8 + 2 x−x^2 [(1,9) Maximum]


3. y=x^2 − 4 x+ 3 [(2,−1) Minimum]


4. y= 3 + 3 x^2 −x^3

[

(0, 3)Minimum,

(2, 7)Maximum

]

5. y= 3 x^2 − 4 x+ 2

[

Minimum at

( 2

3 ,

2

3

)]

6. x=θ( 6 −θ) [Maximum at (3, 9)]


7. y= 4 x^3 + 3 x^2 − 60 x− 12
   [
   Minimum( 2 ,− 88 );
   Maximum(− 2. 5 , 94. 25 )

]

8. y= 5 x−2lnx
   [Minimum at (0.4000, 3.8326)]


9. y= 2 x−ex
   [Maximum at (0.6931,−0.6136)]


10. y=t^3 −

t^2

2

− 2 t+ 4

⎡

⎢

⎣

Minimum at( 1 , 2. 5 );

Maximum at

(

−

2

3

, 4

22

27

)

⎤

⎥

⎦

11. x= 8 t+

1

2 t^2

[Minimum at( 0. 5 , 6 )]

12. Determine the maximum and minimum values
    on the graphy=12cosθ−5sinθin the range
    θ=0toθ= 360 ◦. Sketch the graph over one
    cycle showing relevant points.
    [
    Maximum of 13 at 337. 38 ◦,
    Minimum of−13 at 157. 38 ◦

]

13. Show that the curvey=^23 (t− 1 )^3 + 2 t(t− 2 )
    has a maximum value of^23 and a minimum
    value of−2.

28.4 Practical problems involving

maximum and minimum values

There are manypractical problemsinvolving maxi-

mum and minimum values which occur in science and

engineering. Usually, an equation has to be determined

from given data, and rearranged where necessary, so

that it contains only one variable. Some examples are

demonstrated in Problems 15 to 20.

Problem 15. A rectangular area is formed having

a perimeter of 40cm. Determine the length and

breadth of the rectangle if it is to enclose the

maximum possible area.

Let the dimensions of the rectangle bexandy.Then

the perimeter of the rectangle is (2x+ 2 y). Hence

2 x+ 2 y= 40 ,

or x+y= 20 (1)

Since the rectangle is to enclose the maximum possible

area, a formula for areaAmust be obtained in terms of

one variable only.

AreaA=xy. From equation (1),x= 20 −y

Hence, areaA=( 20 −y)y= 20 y−y^2

dA

dy

= 20 − 2 y= 0

for a turning point, from which,y=10cm

d^2 A

dy^2

=− 2 ,

which is negative, giving a maximum point.

Wheny=10cm,x=10cm, from equation (1).

Hence the length and breadth of the rectangle are

each 10cm, i.e. a square gives the maximum possible

area. When the perimeter of a rectangle is 40cm, the

maximum possible area is 10× 10 =100cm^2.

Problem 16. A rectangular sheet of metal having

dimensions 20cm by 12cm has squares removed

from each of the four corners and the sides bent