Find two positive numbers whose sum is 60 and the product of the square of one number and the other number is maximum.

40 and 20

The correct answer is Option (3) → 40 and 20

Let one number be $x (0 < x <60)$, then the other number is $60-x$ because the sum of two numbers is given to be 60.

Let $f(x)=x^2 (60 - x)$   ...(i)

Differentiating (i) w.r.t. x, we get

$f'(x) = x^2. (-1) + (60 - x). 2x = -3x^2 + 120x$

and $f''(x) = -6x + 120$.

Now $f'(x) = 0⇒-3x^2 + 120x = 0 ⇒-3x (x - 40) = 0$

$⇒x = 0, 40$ but $0 < x < 60$

$⇒x = 40$.

Also $f''(40)=-6 × 40+ 120=-120 < 0$

$⇒ f(x)$ has local maxima at $x = 40$.

Since $x = 40$ is the only extreme point, the local maximum is an absolute maximum.

Hence, the required numbers are 40 and 60 - 40 i.e. 40 and 20.