A sculpture formed from a prism is fixed on a horizontal platform, as shown in the diagram - AQA - A-Level Maths Pure - Question 12 - 2017 - Paper 1

To find the maximum height of the sculpture, we start with the given equation:

$x^2 + 2xy + 2y^2 = 10.$
Using implicit differentiation, we differentiate both sides with respect to $x$:

$2x + 2y + 2x\frac{dy}{dx} + 4y\frac{dy}{dx} = 0.$
Rearranging gives:

$\frac{dy}{dx}(2x + 4y) = -2x - 2y.$
Thus,

$\frac{dy}{dx} = \frac{-2x - 2y}{2x + 4y}.$
We find stationary points by setting $\frac{dy}{dx} = 0$, leading to:

$-2x - 2y = 0 \Rightarrow x + y = 0 \Rightarrow y = -x.$
Now substituting $y = -x$ back into the original equation:

$x^2 + 2x(-x) + 2(-x)^2 = 10 \Rightarrow x^2 - 2x^2 + 2x^2 = 10 \Rightarrow x^2 = 10 \Rightarrow x = \pm\sqrt{10}.$
Substituting these values back to find $y$ gives:

$y = -\sqrt{10} \text{ or } y = \sqrt{10}.$