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

Stationary points occur when
$\frac{dy}{dx} = 0$, therefore:
$2x + 2y = 0 \implies y = -x$.

Substituting $y = -x$ back into the original equation gives:
$x^2 + 2x(-x) + 2(-x)^2 = 10$.

This simplifies to:
$x^2 - 2x^2 + 2x^2 = 10 \implies x^2 = 10.$
Therefore,
$x = \sqrt{10} \text{ or } -\sqrt{10}$.

Using $y = -x$, we find:
$y = -\sqrt{10} \text{ or } \sqrt{10}.$
Thus, the maximum and minimum values of $y$ are $\sqrt{10}$ and $-\sqrt{10}$ respectively.

The maximum value for $y$ is $\sqrt{10}$. Therefore, the maximum vertical height above the platform is:
$Height = 2\sqrt{10} = 6.32 \text{ metres}.$

The height of the sculpture above the platform is thus $6.32$ metres.