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V = PQ \times QR \times \mathrm{height}

= (2n-2x)(n-2x)x

= 2n^2x-6nx^2+4x^3

\frac{dV}{dx}=2n^2 - 12nx +12x^2

Let \frac{dV}{dx}=0

\Rightarrow 6x^2 -6nx + n^2 = 0

x = (\frac{3 \pm \sqrt{3}}{6})n

Since 2x \textless n, ~ x \textless \frac{n}{2} \Rightarrow x = (\frac{3- \sqrt{3}}{6})n

We reject x=\frac{3+\sqrt{3}}{6}n since \frac{3+\sqrt{3}}{6}n > \frac{n}{2}

\therefore Required stationary value of x = \frac{3-\sqrt{3}}{6}n

KS Comments:

Students do not need to carry out the second derivative test since question only ask for the stationary value, instead of maximum of minimum. Question wants students to explain why there is one answer only, so a clear precise explanation will be expected.

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