Page 346 - DMTH201_Basic Mathematics-1

Unit 13: Maxima and Minima

dV Notes
Further, = L 2x 2 4x L 2x 0, for maximum V.
dx
L 2x L 2x 4x = or L 2x L 6x 0

L L
or x = or x
2 6
Second order condition
2
d V 4 L 2x 4 L 2x 8x 8L 24x
dx 2 =

L
= 8L 12L 4L 0, when x ,
2
L
and = 8L 4L 4L 0, when x .
6
L
Thus the volume is largest when x = .
6

L L 2 2L 3
The largest volume = L .
6 3 27

Example: A running track of 440 ft. is to be laid out enclosing a football field, the shape
of which is a rectangle with a semicircle at each end. If the area of the rectangular portion is to be
kept maximum, find the length of its sides.
Solution:
Let x be the length of the rectangular portion and y be the breadth of the football field. Total
perimeter of the running track is
y P 2x 440 2x
P = 2x 2 = 2x + py or y
2
Let A be the area of the rectangular portion.

440x 2x 2
A = x .y

dA 440 4x
Further, = 0 for maximum A
dx
x = 110 ft.
440 220
Also y = 7 70
22
Figure 13.10

Fig. 5.10

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