Higher Engineering Mathematics

306 DIFFERENTIAL CALCULUS

1. y= 3 x^2 − 4 x+ 2

[

Minimum at

( 2

3 ,

2

3

)]

2. x=θ(6−θ) [Maximum at (3, 9)]


3. y= 4 x^3 + 3 x^2 − 60 x− 12
   [
   Minimum (2,−88);
   Maximum(− 2 .5, 94.25)

]

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


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


6. y=t^3 −

t^2

2

− 2 t+ 4

⎡

⎣

Minimum at (1, 2.5);

Maximum at

(

−

2

3

,4

22

27

)

⎤

⎦

7. x= 8 t+

1

2 t^2

[Minimum at (0.5, 6)]

8. Determine the maximum and minimum val-
   ues on the graphy=12 cosθ−5 sinθin the
   rangeθ=0toθ= 360 ◦. Sketch the graph
   over one cycle showing relevant points.
   [
   Maximum of 13 at 337◦ 23 ′,
   Minimum of−13 at 157◦ 23 ′

]

9. 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 max-
imum 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 hav-

ing a perimeter of 40 cm. 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 pos-

sible 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=10 cm

d^2 A

dy^2

=−2,

which is negative, giving a maximum point.

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

Hence the length and breadth of the rectangle

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

possible area. When the perimeter of a rectangle

is 40 cm, the maximum possible area is 10× 10 =

100 cm^2.

Problem 16. A rectangular sheet of metal hav-

ing dimensions 20 cm by 12 cm has squares

removed from each of the four corners and

the sides bent upwards to form an open box.

Determine the maximum possible volume of

the box.

The squares to be removed from each corner are

shown in Fig. 28.8, having sidesxcm. When the

sides are bent upwards the dimensions of the box

will be:

Figure 28.8

length (20− 2 x) cm, breadth (12− 2 x) cm and

height,xcm.

Volume of box,

V=(20− 2 x)(12− 2 x)(x)

= 240 x− 64 x^2 + 4 x^3