Question
Find two positive numbers x and y such that x + y = 60 and xy3 is maximum.
Solution
Here x + y = 60 ⇒ y = 60 – x ...(1)
Let f (x) = x y3 = x (60 – x)3 [∵ of (1)]
Let f (x) = x y3 = x (60 – x)3 [∵ of (1)]
= (60 – x)2 [ – 3 x + 60 – x] = (60 – x)2 (– 4 x + 60)
= 4 (60 – x)2 (15 – x)
f '(x) = 0 ⇒ 4 (60 – x)2 (15 – x) = 0 ⇒ x = 15, 60
Rejecting a = 60 as 0 < x < 60, we get, x = 15
When x < 15 slightly, f ' (x) = 4 (+) (+) = + ve
When x > 15 slightly, f ' (a) = 4 (+) (–) = – ve
∴ at x = 15, f ' (x) changes from + ve to – ve
∴ f (x) has local maximum at x = 15.
But f (x) has only one extreme point x = 15
∴ f (x) is maximum when x = 15, y = 60 – 15 = 45
∴ x = 15, y = 45
