If x and y are positive numbers such that x + y = 20 and the product…

The correct answer is Option (4) → \(x = 5,\ y = 15\)

Since x = 20 − y, with 0 < y < 20, the product is

    \(P = (20 - y)\,y^3\)
    \(P = 20y^3 - y^4\)

Setting the derivative equal to zero:

    \( \displaystyle \frac{dP}{dy} = 60y^2 - 4y^3 \)

    \( = 4y^2(15 - y) \)

Since y > 0, the only critical point is y = 15. The second derivative test:

    \( \displaystyle \frac{d^2P}{dy^2} = 120y - 12y^2 \)

    \(\text{at } y = 15:\ 1800 - 2700 = -900 \lt 0\)

So P is a maximum at y = 15, and then x = 20 − 15 = 5.

Why the other options are incorrect

• Option (1) \(x = 15,\ y = 5\) — Incorrect. The values are swapped: \(xy^3 = 15 \times 125 = 1875\), far below \(5 \times 3375 = 16875\).

• Option (2) \(x = 10,\ y = 10\) — Incorrect. Equal numbers maximise xy, not \(xy^3\); here \(xy^3 = 10000\).

• Option (3) \(x = 4,\ y = 16\) — Incorrect. \(xy^3 = 4 \times 4096 = 16384\), less than 16875.