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 vertical height of the sculpture, we need to differentiate the given equation with respect to x, and apply implicit differentiation.

1. Start with the equation of the cross-section: $x^2 + 2xy + 2y^2 = 10$

2. Differentiate both sides implicitly: rac{d}{dx}(x^2) + rac{d}{dx}(2xy) + rac{d}{dx}(2y^2) = 0 This results in: 2x + 2y + 2x rac{dy}{dx} + 4y rac{dy}{dx} = 0

3. Combine like terms: 2x + 2y + (2x + 4y)rac{dy}{dx} = 0

4. Set the derivative to zero to find stationary points:

   $$$$

ightarrow rac{dy}{dx} = -rac{2x + 2y}{2x + 4y}$$ Setting this equal to zero gives:

ightarrow x = -y$$ 5. Substitute $x = -y$ back into the original equation: $$(-y)^2 + 2(-y)y + 2y^2 = 10$$ This simplifies to: $$y^2 - 2y^2 + 2y^2 = 10 ightarrow y^2 = 10$$ Thus, we get: $$y = rac{ ext{±} oot{10}}{ ext{±1}}$$ 6. The maximum positive value is: $$y = rac{ ext{±} oot{10}}{ ext{+}} = ext{±3.16}$$, which means the maximum vertical height would be 6.32 m when accounting for vertical positioning. Therefore, the maximum height above the platform of the sculpture is approximately 6.32 m.