2026-08-31T11:35:12.809715 image/svg+xml Matplotlib v3.10.9, https://matplotlib.org/ © EAGLE BEACON GLOBAL Figure 3 shows a tank in the shape of a cuboid wi...

This is a standard optimisation chain: use the fixed volume to eliminate one variable from the surface area, differentiate, solve for the stationary point, and test its nature with the second derivative. The reason a constraint is needed is that \(S\) depends on both \(x\) and \(h\), and calculus in one variable requires a function of \(x\) alone.

(a) Showing that \(S = 2x^{2} + \dfrac{108}{x}\) [4 marks]

The tank is a cuboid of base \(2x\) by \(x\) and height \(h\), so its volume is the product of the three dimensions. The given volume of \(36\,\mathrm{m}^{3}\) is the constraint:

\[2x \times x \times h = 2x^{2}h = 36 \quad \Longrightarrow \quad h = \frac{36}{2x^{2}} = \frac{18}{x^{2}} \qquad \textbf{M1 A1}\]

The M1 is for forming a correct volume equation and the A1 for rearranging it to give \(h\) in terms of \(x\). Now the surface area. The top is open, so there is one base of area \(2x \times x = 2x^{2}\), plus four vertical faces: two of them measure \(2x\) by \(h\) and two measure \(x\) by \(h\):

\[S = 2x^{2} + 2\left(2xh\right) + 2\left(xh\right) = 2x^{2} + 6xh \qquad \textbf{M1}\]

Substituting the expression for \(h\) eliminates it:

\[S = 2x^{2} + 6x \times \frac{18}{x^{2}} = 2x^{2} + \frac{108}{x} \qquad \textbf{A1}\]

as required. Since the result is printed, the marks belong to the derivation. Two errors are frequent. The first is including a top, which adds a second \(2x^{2}\) and gives \(4x^{2} + \dfrac{108}{x}\); the phrase "open top" and the request for the area "of the base and the four vertical faces" both rule it out. The second is counting only two vertical faces, or taking all four as \(2x\) by \(h\), which gives \(8xh\) instead of \(6xh\). Check the units: \(2x^{2}\) and \(\dfrac{108}{x}\) are both areas in \(\mathrm{m}^{2}\) because the \(108\) carries units of \(\mathrm{m}^{3}\).

(b) The first and second derivatives [3 marks]

Rewrite the fraction as a negative power so the power rule applies:

\[S = 2x^{2} + 108x^{-1}\] \[\frac{\mathrm{d}S}{\mathrm{d}x} = 4x - 108x^{-2} = 4x - \frac{108}{x^{2}} \qquad \textbf{M1 A1}\] \[\frac{\mathrm{d}^{2}S}{\mathrm{d}x^{2}} = 4 + 216x^{-3} = 4 + \frac{216}{x^{3}} \qquad \textbf{A1}\]

The M1 is for differentiating a negative power at all and survives a coefficient slip; the two A1 marks require each derivative to be exactly right. Watch the signs: differentiating \(108x^{-1}\) gives \(-108x^{-2}\), and differentiating that in turn gives \(+216x^{-3}\), because multiplying by the index \(-2\) turns the negative coefficient positive. Losing that second sign change is the usual route to a wrong conclusion in part (d).

(c) The stationary value [4 marks]

A stationary value occurs where the first derivative is zero, which is the definition to quote:

\[4x - \frac{108}{x^{2}} = 0 \qquad \textbf{M1}\]

Multiply through by \(x^{2}\), legitimate because \(x \gt 0\) for a physical tank:

\[4x^{3} = 108 \quad \Longrightarrow \quad x^{3} = 27 \qquad \textbf{M1}\] \[x = 3 \qquad \textbf{A1}\]

Only the real cube root is relevant, and \(x = 3\) is the value the question states, so this part is a "show that" as far as \(x\) is concerned. Substituting back into the expression for \(S\), not into a derivative:

\[S = 2\left(3\right)^{2} + \frac{108}{3} = 18 + 36 = 54 \qquad \textbf{A1}\]

so the stationary value is \(S = 54\,\mathrm{m}^{2}\). The commonest loss is substituting \(x = 3\) into \(\dfrac{\mathrm{d}S}{\mathrm{d}x}\) and reporting zero, which answers a different question.

(d) Maximum or minimum [2 marks]

Evaluate the second derivative at the stationary point:

\[\left.\frac{\mathrm{d}^{2}S}{\mathrm{d}x^{2}}\right|_{x=3} = 4 + \frac{216}{27} = 4 + 8 = 12 \qquad \textbf{M1}\] \[12 \gt 0, \quad \text{so } S = 54 \text{ is a minimum} \qquad \textbf{A1}\]

The M1 is for substituting into the second derivative, the A1 for the correct conclusion drawn from its sign. A positive second derivative means the gradient is increasing, so the curve is concave upwards and the stationary point is a minimum. The A1 requires the sign to be stated as the reason; simply asserting "minimum" does not show the test was used. The answer also makes physical sense: with a fixed volume, a very thin tall tank and a very wide flat tank both have huge surface area, so the stationary value between them must be the least material, never the most.

(e) The height [1 mark]

Use the constraint from part (a) with \(x = 3\):

\[h = \frac{18}{3^{2}} = \frac{18}{9} = 2\,\mathrm{m} \qquad \textbf{B1}\]

This is a B mark for the correct value, and the unit should be given. Check the whole answer against both original conditions: the volume is \(2\left(3\right)\left(3\right)\left(2\right) = 36\,\mathrm{m}^{3}\), and the surface area is \(2\left(9\right) + 6\left(3\right)\left(2\right) = 18 + 36 = 54\,\mathrm{m}^{2}\). Both match, so the optimisation is confirmed.

Examination takeaway. In an optimisation question, use the fixed quantity to express the redundant variable in terms of the one you will differentiate, and read the surface description carefully, since an open top removes a face. Substitute the optimal value into the original function for the stationary value, and always quote the sign of the second derivative as the reason for calling it a maximum or a minimum.