If 2x+y=6 then the maximum value of x^2y is:

The correct answer is Option (3) → 8

As usual in such questions, x and y are taken as positive numbers (for large negative x the product \(x^2y\) has no upper limit). From the given condition, y = 6 − 2x, so let

    \(P = x^2(6-2x)\)
    \(P = 6x^2 - 2x^3\)

Setting the first derivative equal to zero:

    \( \displaystyle \frac{dP}{dx} = 12x - 6x^2 \)

    \(6x(2 - x) = 0\)
    \(x = 2 \quad (x \gt 0)\)

The second derivative test:

    \( \displaystyle \frac{d^2P}{dx^2} = 12 - 12x \)

    \(\text{at } x = 2: \ 12 - 24 = -12 \lt 0\)

So P is maximum at x = 2, where y = 6 − 4 = 2:

    \(x^2y = 4 \times 2 = 8\)