UPSC Maths 2025 Paper 1 Q1c — Solution

10 marks · Section A

Question

A rectangular sheet of metal of length 6 meters and width 2 meters is given. Four equal squares are removed from the four corners. The sides of this sheet are now folded up to form an open rectangular box. Find approximately the height of the box, such that the volume of the box is maximum.

Technique

Express the volume as a single-variable function of the cut size, then maximise it on the feasible domain $0 < x < 1$ using the first and second derivative tests.

Solution

Step 1 — Set up the volume function.

Let $x$ be the side of each removed square, which equals the height of the folded box. After cutting and folding:

length of base $= 6 - 2x$ ,
width of base $= 2 - 2x$ ,
height $= x$ .

$V(x) = x(6-2x)(2-2x), \qquad 0 < x < 1$ (the domain requires $x<1$ so that the width $2-2x>0$ ).

Expand: $V(x) = x(12 - 12x - 4x + 4x^2) = 4x^3 - 16x^2 + 12x.$

Step 2 — Critical points.

$V'(x) = 12x^2 - 32x + 12 = 4(3x^2 - 8x + 3).$ Set $V'(x)=0$ : $x = \frac{8 \pm \sqrt{64 - 36}}{6} = \frac{8 \pm \sqrt{28}}{6} = \frac{4 \pm \sqrt 7}{3}.$

The two roots are $x = \dfrac{4-\sqrt7}{3}\approx 0.4514$ and $x = \dfrac{4+\sqrt7}{3}\approx 2.2153$ . Only the first lies in the feasible interval $(0,1)$ .

Step 3 — Second derivative test.

$V''(x) = 24x - 32.$ At $x=\dfrac{4-\sqrt7}{3}$ : $V'' = 24\cdot\dfrac{4-\sqrt7}{3} - 32 = 8(4-\sqrt7) - 32 = -8\sqrt7 < 0,$ so this is a maximum.

Step 4 — Value.

$x = \frac{4-\sqrt7}{3} \approx 0.451 \text{ m}, \qquad V_{\max} \approx 2.525 \text{ m}^3.$

Answer

$\boxed{\;\text{Height } x = \dfrac{4-\sqrt7}{3} \approx 0.45 \text{ m (maximum volume } \approx 2.52\ \text{m}^3).\;}$