Find two positive numbers whose sum is 60 and the product of the squa…

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.