2026-08-31T11:34:02.160849 image/svg+xml Matplotlib v3.10.9, https://matplotlib.org/ © EAGLE BEACON GLOBAL Figure 2 shows a solid cuboid with a square base ...

This is a constrained optimisation problem, and the standard shape of the solution is always the same: two variables, one constraint, so use the constraint to eliminate one variable and reduce the target quantity to a function of a single variable. Only then can calculus be applied. Here the fixed surface area is the constraint and the volume is the target, so \(h\) must be written in terms of \(x\) and substituted into \(V = x^2 h\).

(a) Show that \(V = 75x - \tfrac{1}{2}x^3\) [3]

The result is printed, so the marks are for the derivation. A cuboid with a square base of side \(x\) and height \(h\) has two square faces of area \(x^2\) and four rectangular faces of area \(xh\), so the total surface area is

\[2x^2 + 4xh = 300 \qquad \textbf{[M1]}\]

The [M1] is for a correct surface-area expression. The most frequent error is \(2x^2 + 4x^2h\) or using only two rectangular faces; count the faces deliberately, since the base is square and closed on both ends.

Rearranging for \(h\):

\[h = \frac{300 - 2x^2}{4x} = \frac{150 - x^2}{2x} \qquad \textbf{[M1]}\]

The second [M1] is for making \(h\) the subject. Now substitute into the volume:

\[V = x^2 h = x^2 \times \frac{150 - x^2}{2x} = \frac{x\left(150 - x^2\right)}{2} = 75x - \frac{1}{2}x^3 \qquad \textbf{[A1]}\]

The [A1] is for reaching the printed form with the algebra shown. One \(x\) cancels between \(x^2\) and the \(2x\) in the denominator; cancelling both and writing \(V = (150 - x^2)/2\) loses the accuracy mark.

(b) Find \(\dfrac{\mathrm{d}V}{\mathrm{d}x}\) and show \(V\) is stationary when \(x = 5\sqrt{2}\) [3]

Differentiate term by term, multiplying by the power and reducing it by one:

\[\frac{\mathrm{d}V}{\mathrm{d}x} = 75 - \frac{3}{2}x^2 \qquad \textbf{[M1 A1]}\]

The [M1] is for the attempt at differentiation, evidenced by at least one term correct; the [A1] is for both terms. Note that \(\tfrac{1}{2} \times 3 = \tfrac{3}{2}\), and the derivative of the linear term \(75x\) is the constant \(75\), not \(75x\).

A stationary value occurs where the derivative is zero:

\[75 - \frac{3}{2}x^2 = 0 \quad\Rightarrow\quad \frac{3}{2}x^2 = 75 \quad\Rightarrow\quad x^2 = 50 \quad\Rightarrow\quad x = 5\sqrt{2}\]

taking the positive root because \(x\) is a length, so \(x \gt 0\). That final [A1] requires the rejection of \(x = -5\sqrt{2}\) to be visible or at least implied by taking the positive value. \(\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}\), which is the printed form; leaving \(\sqrt{50}\) does not match the target of a "show that".

(c) Maximum volume in the form \(k\sqrt{2}\) [2]

Substitute \(x = 5\sqrt{2}\) into \(V\). The cube is the step where marks are lost, so do it carefully:

\[\left(5\sqrt{2}\right)^3 = 5^3 \times \left(\sqrt{2}\right)^3 = 125 \times 2\sqrt{2} = 250\sqrt{2}\] \[V = 75\left(5\sqrt{2}\right) - \frac{1}{2}\left(250\sqrt{2}\right) = 375\sqrt{2} - 125\sqrt{2} \qquad \textbf{[M1]}\] \[V = 250\sqrt{2}\,\mathrm{cm}^3 \qquad \textbf{[A1]}\]

The [M1] is for the substitution, so it survives an arithmetic slip in the cube; the [A1] needs the exact value \(k = 250\). Because the question specifies the form \(k\sqrt{2}\) with \(k\) an integer, a decimal such as \(353.6\) scores nothing for the accuracy mark, however correct it is numerically.

(d) Justify that this value is a maximum [1]

Differentiate a second time:

\[\frac{\mathrm{d}^2V}{\mathrm{d}x^2} = -3x\]

At \(x = 5\sqrt{2}\) this equals \(-15\sqrt{2}\), which is negative, so the curve is concave there and the stationary value is a maximum [B1].

The mark is for a genuine justification, not an assertion. Two things must appear: the second derivative evaluated at the stationary value, and the conclusion drawn from its sign being negative. Writing "it is a maximum because the volume cannot get any bigger" earns nothing. An acceptable alternative is a sign test on \(\mathrm{d}V/\mathrm{d}x\) either side of \(5\sqrt{2} \approx 7.07\): at \(x = 7\), \(75 - \tfrac{3}{2}(49) = 1.5 \gt 0\), and at \(x = 7.5\), \(75 - \tfrac{3}{2}(56.25) = -9.375 \lt 0\), so the gradient changes from positive to negative, confirming a maximum.

Check: the answer should be physically sensible. With \(x = 5\sqrt{2} \approx 7.07\,\mathrm{cm}\), the height is \(h = \dfrac{150 - 50}{2 \times 5\sqrt{2}} = \dfrac{100}{10\sqrt{2}} = 5\sqrt{2}\,\mathrm{cm}\), so the optimal cuboid is in fact a cube. Its surface area is \(6\left(5\sqrt{2}\right)^2 = 6 \times 50 = 300\,\mathrm{cm}^2\), matching the constraint, and its volume is \(\left(5\sqrt{2}\right)^3 = 250\sqrt{2}\,\mathrm{cm}^3\), matching part (c).