Q.An open topped box is to be constructed by removing equal squares from each corner of a $3$ metre by $8$ metre rectangular sheet of aluminium and folding up the sides. Find the volume of the largest such box.

The key idea is to express the volume as a function of the side length $x$ of the cut-out squares, then maximise it using differentiation.

Step 1: Write the volume function.

After cutting squares of side $x$ from each corner and folding, the box has:

• length $= 8 - 2x$,
• width $= 3 - 2x$,
• height $= x$.

Hence the volume is

Step 2: Find the critical point.

Differentiate: $V'(x) = 12x^2 - 44x + 24$. Set $V'(x) = 0$ and divide by 4:

Solving gives $x = \frac{2}{3}$ or $x = 3$. Since $x$ must be less than half the smaller side ($3/2 = 1.5$), discard $x = 3$. So $x = \frac{2}{3}$ is the only feasible critical point.

Step 3: Confirm maximum and compute volume. …

This is a classic optimisation problem from calculus — you have a fixed rectangular sheet, you cut identical squares of side $x$ from each corner, fold up the flaps, and get an open-top box. The question asks: what size square gives the largest possible volume?

The key insight is that the box's dimensions are completely determined by $x$. The original sheet is $3$ m by $8$ m. After cutting $x \times x$ squares from each corner, the base of the box becomes a rectangle of length $8 - 2x$ and width $3 - 2x$. The height of the box is exactly $x$, the side of the cut-out square. So the volume is simply:

We want the value of $x$ that maximises $V(x)$, but $x$ cannot be any number — it must be positive and small enough that the base dimensions stay positive. That gives the feasible domain: $0 < x < 1.5$ (since $3 - 2x > 0$). Within this interval, $V(x)$ is a smooth cubic, and its maximum occurs either at a critical point (where $V'(x)=0$) or at an endpoint. The endpoints give $V=0$, so the maximum is interior.

Let's work through it step by step.

1. Write the volume function and simplify.

Multiply the two linear factors first:

Then multiply by $x$:

2. Differentiate to find critical points.

Set $V'(x) = 0$:

Divide through by 4 to simplify:

3. Solve the quadratic.

Using the quadratic formula:

So the two roots are:

4. Check which root lies in the feasible domain.

   $x = 3$ is outside $0 < x < 1.5$, so it is not physically possible. The only feasible critical point is $x = \frac{2}{3}$ metre.

5. Confirm it gives a maximum.

   You can use the second derivative test. Compute $V''(x) = -44 + 24x$. At $x = \frac{2}{3}$:

   $V''\left(\frac{2}{3}\right) = -44 + 24 \cdot \frac{2}{3} = -44 + 16 = -28 < 0$ …

Method: Optimization from a Word Problem (Objective–Constraint–Reduce–Differentiate)

The general six-step framework for turning any applied maximize/minimize word problem into a calculus problem.

Steps

Step 1: Name the quantity to optimize and express it with variables

Identify what the problem wants made as large or small as possible (here, volume), and write a formula for it using however many variables the geometry naturally involves.

Step 2: Identify the constraint linking those variables

Look for the fixed real-world limitation stated in the problem (fixed sheet dimensions, fixed perimeter, fixed surface area, etc.) and write it as a relation between the variables.

Step 3: Reduce the objective to a single variable

Use the constraint (or the geometry directly, as with dimensions expressed in terms of one cut length) to rewrite the objective function so it depends on only one variable.

Step 4: Differentiate, solve $Q'(x)=0$, and restrict to the physically valid domain …

Common Mistakes

Mistake 1: Reducing each side by $x$ instead of $2x$

Why it's wrong: cutting a square of side $x$ from each corner removes a strip of width $x$ from both ends of each side of the rectangle, so each full dimension shrinks by $2x$, not $x$. Writing the base dimensions as $(3-x)$ and $(8-x)$ instead of $(3-2x)$ and $(8-2x)$ gives an entirely wrong volume function. Correct approach: picture folding up all four flaps — each side loses one square-width from each of its two ends, so the total reduction per dimension is $2x$.

Mistake 2: Accepting a critical point without checking the physical domain …