The two positive numbers whose sum is 40 and the product of the cube…

The correct answer is Option (3) → 30, 10

Let the numbers be x and 40 − x, with 0 < x < 40, and let x be the number that is cubed.

    \( \displaystyle P(x)=x^{3}(40-x)=40x^{3}-x^{4} \)

    \( \displaystyle P'(x)=120x^{2}-4x^{3}=4x^{2}(30-x) \)

For 0 < x < 40, P′(x) = 0 only at x = 30.

    \( \displaystyle P''(x)=240x-12x^{2} \)
    \( \displaystyle P''(30)=7200-10800=-3600\lt 0 \)

So P is maximum at x = 30, and the numbers are 30 and 40 − 30 = 10.

Shortcut: Check the products in the options, cubing the larger number:

    \( \displaystyle 15,\ 25:\ 25^{3}\times 15=234375 \)
    \( \displaystyle 5,\ 35:\ 35^{3}\times 5=214375 \)
    \( \displaystyle 30,\ 10:\ 30^{3}\times 10=270000 \)
    \( \displaystyle 12,\ 28:\ 28^{3}\times 12=263424 \)

The largest is 270000, for 30 and 10.

Tip: To maximise \( x^{m}y^{n} \) with x + y fixed, divide the sum in the ratio m : n. Here 3 : 1 splits 40 into 30 and 10.