Application of derivatives — rate of change, tangent/normal, maxima/minima strategy

Question

A rectangular sheet of metal has dimensions $48 \text{ cm} \times 36 \text{ cm}$ . Equal squares of side $x$ are cut from each corner and the flaps are folded up to form an open box. Find the value of $x$ that maximises the volume of the box.

(CBSE 12 Board / JEE Main pattern)

Which Application? A Decision Flowchart

Before solving, we need to recognise what type of derivative application we are dealing with. Here is the strategy:

flowchart TD
    A[Derivative Application Problem] --> B{What is asked?}
    B -->|Rate of something changing| C[Rate of Change]
    B -->|Equation of tangent/normal| D[Tangent & Normal]
    B -->|Maximum or minimum value| E[Maxima / Minima]
    C --> C1[Identify variable & time relation]
    C1 --> C2["Differentiate w.r.t. time: dy/dt"]
    D --> D1["Find dy/dx at given point"]
    D1 --> D2["Tangent: y - y1 = m(x - x1)"]
    D1 --> D3["Normal: slope = -1/m"]
    E --> E1["Express quantity in one variable"]
    E1 --> E2["Set f'(x) = 0, find critical points"]
    E2 --> E3["Use second derivative test"]

Solution — Step by Step

After cutting squares of side $x$ from each corner and folding:

Length of box = $48 - 2x$
Width of box = $36 - 2x$
Height of box = $x$

V(x) = x(48 - 2x)(36 - 2x)

Expanding: $V(x) = x(1728 - 96x - 72x + 4x^2) = 4x^3 - 168x^2 + 1728x$

V'(x) = 12x^2 - 336x + 1728

Setting $V'(x) = 0$ :

12x^2 - 336x + 1728 = 0

x^2 - 28x + 144 = 0

Using the quadratic formula: $x = \dfrac{28 \pm \sqrt{784 - 576}}{2} = \dfrac{28 \pm \sqrt{208}}{2}$

$x = \dfrac{28 \pm 4\sqrt{13}}{2} = 14 \pm 2\sqrt{13}$

So $x \approx 14 + 7.21 = 21.21$ or $x \approx 14 - 7.21 = 6.79$

Since the width is 36 cm, we need $36 - 2x > 0$ , giving $x < 18$ . So $x \approx 21.21$ is rejected.

For $x = 14 - 2\sqrt{13}$ :

V''(x) = 24x - 336

V''(14 - 2\sqrt{13}) = 24(14 - 2\sqrt{13}) - 336 = -48\sqrt{13} < 0

Since $V'' < 0$ , this is a maximum.

\boxed{x = 14 - 2\sqrt{13} \approx 6.79 \text{ cm}}

Why This Works

Optimization problems follow a fixed recipe: express the quantity to maximise (or minimise) as a function of a single variable, differentiate, find critical points, and test them. The second derivative test tells us whether a critical point is a maximum ( $f'' < 0$ ) or minimum ( $f'' > 0$ ).

The constraint (dimensions must be positive) eliminates one root automatically. This is typical of CBSE/JEE problems where physical constraints narrow down the answer.

Alternative Method — First Derivative Test

Instead of computing $V''(x)$ , check the sign of $V'(x)$ around $x = 14 - 2\sqrt{13}$ :

For $x$ slightly less than $6.79$ : $V'(x) > 0$ (increasing)
For $x$ slightly more than $6.79$ : $V'(x) < 0$ (decreasing)

Sign changes from $+$ to $-$ , confirming a local maximum. This method is more reliable when $V''(x) = 0$ at the critical point (which happens in some tricky JEE problems).

In JEE Main, about 70% of Application of Derivatives questions are maxima/minima. The remaining split between rate of change and tangent/normal. Always start by classifying the problem type — it saves you from applying the wrong technique.

Common Mistake

Students often forget to check the domain constraint. In this problem, both $x > 0$ and $48 - 2x > 0$ and $36 - 2x > 0$ must hold simultaneously, giving $0 < x < 18$ . Picking the root outside this range gives a meaningless answer — negative dimensions.

Another frequent slip: forgetting the factor of 2 in " $48 - 2x$ ". Each corner loses a square from both sides, so the total reduction in length is $2x$ , not $x$ .