Question

1. maximum mark: 8

the diagram shows a square prism of length ( l ) cm and cross - section ( x ) cm by ( x ) cm. given that the surface area of the prism is ( k ) ( \text{cm}^2 ), where ( k ) is a constant.
(a) show that ( l=\frac{k - 2x^{2}}{4x} ), 2
(b) use calculus to prove that the maximum volume of the prism occurs when it is a cube. 6

Explanation:

Step1: Write surface area formula

The prism has 2 square faces (area $x^2$ each) and 4 rectangular faces (area $xl$ each). So total surface area:
$$k = 2x^2 + 4xl$$

Step2: Rearrange to solve for $l$

Isolate the term with $l$ first:
$$4xl = k - 2x^2$$
Divide both sides by $4x$:
$$l = \frac{k - 2x^2}{4x}$$

Step3: Define volume of the prism

Volume $V$ is cross-sectional area times length:
$$V = x^2 l$$
Substitute $l = \frac{k - 2x^2}{4x}$ into the volume formula:
$$V = x^2 \cdot \frac{k - 2x^2}{4x} = \frac{x(k - 2x^2)}{4} = \frac{kx - 2x^3}{4}$$

Step4: Find first derivative of $V$

Differentiate $V$ with respect to $x$ using power rule:
$$\frac{dV}{dx} = \frac{k - 6x^2}{4}$$

Step5: Find critical points

Set $\frac{dV}{dx}=0$ and solve for $x$:
$$\frac{k - 6x^2}{4} = 0$$
$$k - 6x^2 = 0$$
$$x^2 = \frac{k}{6} \implies x = \sqrt{\frac{k}{6}}$$

Step6: Verify it is a maximum

Find the second derivative of $V$:
$$\frac{d^2V}{dx^2} = \frac{-12x}{4} = -3x$$
Since $x>0$, $\frac{d^2V}{dx^2} < 0$, so this critical point is a maximum.

Step7: Show prism is a cube

Substitute $x = \sqrt{\frac{k}{6}}$ into $l = \frac{k - 2x^2}{4x}$:
First calculate $2x^2 = 2 \cdot \frac{k}{6} = \frac{k}{3}$
$$l = \frac{k - \frac{k}{3}}{4\sqrt{\frac{k}{6}}} = \frac{\frac{2k}{3}}{4\sqrt{\frac{k}{6}}} = \frac{2k}{12\sqrt{\frac{k}{6}}} = \frac{k}{6} \cdot \sqrt{\frac{6}{k}} = \sqrt{\frac{k}{6}}$$
Since $l = x$, all edges are equal, so the prism is a cube.

Answer:

(a) Derived as shown: $l = \frac{k - 2x^2}{4x}$
(b) Proven that the maximum volume occurs when the prism is a cube.