The curve $y = 15 - x^2$ and the isosceles triangle OPQ are shown on the diagram below - AQA - A-Level Maths Pure - Question 7 - 2022 - Paper 2

To find the maximum area of triangle OPQ, we take the derivative of the area function:

$A = 15q - q^3$

Differentiating with respect to $q$ gives:

$\frac{dA}{dq} = 15 - 3q^2$

Setting the derivative to zero to find the critical points:

$15 - 3q^2 = 0 \implies q^2 = 5 \implies q = \sqrt{5}$

Next, we confirm that this is a local maximum by checking the second derivative:

$\frac{d^2A}{dq^2} = -6q$

Evaluating at $q = \sqrt{5}$ gives:

$\frac{d^2A}{dq^2} \bigg|_{q = \sqrt{5}} = -6\sqrt{5} < 0$

Thus, we have a local maximum. To find the exact maximum area, substitute $q = \sqrt{5}$ back into the area formula:

$A = 15\sqrt{5} - (\sqrt{5})^3 = 15\sqrt{5} - 5\sqrt{5} = 10\sqrt{5}$

Thus, the exact maximum area of triangle OPQ is $10\sqrt{5}$.