The sum of two positive numbers is 9. The maximum value of the product of the first number and the square of the second number is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 108
Let the second number be y, so the first is 9 − y, with 0 < y < 9. The product is
\(P = (9 - y)\,y^2\)
\(P = 9y^2 - y^3\)
Differentiating:
\( \displaystyle \frac{dP}{dy} = 18y - 3y^2 \)
\( = 3y(6 - y) \)
Since y > 0, the only critical point is y = 6. The second derivative test:
\( \displaystyle \frac{d^2P}{dy^2} = 18 - 6y \)
\(\text{at } y = 6:\ 18 - 36 = -18 \lt 0\)
so P is a maximum at y = 6. Then the first number is 9 − 6 = 3 and
\(P = 3 \times 6^2 = 108\)