Question

Two positive numbers x and y such that x+y=60 and xy^3 is maximum. prove it

Answer 1

Solution :

The two numbers are x and y such that x + y = 60.

y = 60 - x

Let f(x) = xy³

f(x) = x(60 - x)³

f'(x) = (60 - x)² - 3x(60 - x)²

= (60 - x)² [60- x - 3x]

= (60 - x) (60 - 4x)

And.f"(x)= -2(60 - x)(60 - 4x) - 4 (60 - x)²

= -2(60 - x)[ 60 - 4x + 2(60 - x)]

= -2(60 - x)(180 - 6x)

= -12(60-x)(30-x)

Now, f"(x) =0

x= 60 or x =15

When x = 60,f(x) = 0.

When x = 15, f"(x) = -12(60 - 15)(30 - 15) = -12 x 45 x 15 < 0.

By second derivative test, x = 15 is a point of local maxima of f. Thus, function xy³ is maximum when x = 15 andy = 60 - 15 = 45.

Hence, the required numbers are 15 and 45.