Application of Derivatives: step-by-step worked examples for JEE+CBSE

Application of Derivatives: Step-by-Step Worked Examples (6)

Question

A box without a top is to be made from a square sheet of side $24$ cm by cutting equal squares of side $x$ from each corner and folding up the flaps. Find the value of $x$ that maximizes the volume, and compute the maximum volume.

Solution — Step by Step

After cutting and folding: base is a square of side $(24 - 2x)$ , height is $x$ . So:

V(x) = x(24 - 2x)^2, \quad 0 < x < 12

Expand: $V = x(576 - 96x + 4x^2) = 4x^3 - 96x^2 + 576x$ .

V'(x) = 12x^2 - 192x + 576 = 12(x^2 - 16x + 48)

Setting $V'(x) = 0$ : $x^2 - 16x + 48 = 0$ , so $x = \dfrac{16 \pm \sqrt{256 - 192}}{2} = \dfrac{16 \pm 8}{2}$ , giving $x = 12$ or $x = 4$ .

$x = 12$ would mean cutting away the entire sheet (base side $= 0$ ), so it’s a boundary minimum, not a maximum. Take $x = 4$ .

$V''(x) = 24x - 192$ . At $x = 4$ : $V''(4) = 96 - 192 = -96 < 0$ . Concave down, so $x = 4$ is a local maximum. ✓

V_{\max} = 4(24 - 8)^2 = 4 \times 256 = 1024 \text{ cm}^3

Final Answer: $x = 4$ cm gives maximum volume $1024$ cm $^3$ .

Why This Works

Optimization reduces a geometric question to a one-variable calculus problem. The first derivative locates critical points; the second derivative (or sign analysis around them) tells us which is a max and which is a min. The domain restrictions $0 < x < 12$ rule out unphysical solutions.

A neat sanity check: at the optimal $x = 4$ , the cut-square side is one-sixth of the original sheet side. This $1:6$ ratio is the universal answer for any square sheet — try it with a sheet of side $30$ and you’ll get $x = 5$ , again one-sixth.

Alternative Method

Use AM-GM on $V = x(24 - 2x)^2 = \dfrac{1}{4} \cdot 4x \cdot (24-2x)(24-2x)$ . With three terms summing to $4x + 24 - 2x + 24 - 2x = 48$ , AM-GM gives max product when $4x = 24 - 2x$ , i.e., $x = 4$ . No calculus needed.

Forgetting the domain constraint $0 < x < 12$ leads students to accept $x = 12$ as a valid critical point. Always check whether your critical points are inside the feasible region for the geometric problem.

CBSE Class 12 boards ask this exact problem with different sheet dimensions almost every other year. The setup is identical; only the numbers change. Drill the recipe.

Question

Solution — Step by Step

Why This Works

Alternative Method

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