Q.A rectangular sheet of tin $45$ cm by $24$ cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum?

Concept: Optimization Word Problem (Maximising volume by cutting squares from a rectangular sheet).

Let the side of each square cut off be $x$ cm. After cutting and folding, the box dimensions are:

• Length: $45 - 2x$
• Width: $24 - 2x$
• Height: $x$

Volume $V(x) = x(45 - 2x)(24 - 2x) = x(1080 - 138x + 4x^2) = 4x^3 - 138x^2 + 1080x$.

Differentiate and set to zero:

$V'(x) = 12x^2 - 276x + 1080 = 0 \implies x^2 - 23x + 90 = 0$. …

This is a classic optimization problem from calculus — you’re given a flat rectangular sheet and you cut identical squares from each corner, then fold up the flaps to form an open box. The goal is to choose the side length of those squares so that the box’s volume is as large as possible.

The key insight: the box’s dimensions depend directly on the cut size. If you cut a square of side $x$ from each corner, the length and width of the base each shrink by $2x$ (one $x$ from each side), and the height of the box becomes exactly $x$. So volume = (length of base) × (width of base) × (height). That gives a function of $x$, and we find its maximum using differentiation — the standard calculus approach for such “maximize a quantity” problems.

Let’s work through it step by step.

1. Define the variable and the volume function. Let $x$ cm be the side length of the square cut from each corner. Original sheet: length $45$ cm, width $24$ cm. After cutting and folding:
   • Length of base = $45 - 2x$
   • Width of base = $24 - 2x$
   • Height of box = $x$ Volume $V$ is:

2. Determine the feasible domain of $x$.

   Since dimensions must be positive:

   $45 - 2x > 0 \Rightarrow x < 22.5$

   $24 - 2x > 0 \Rightarrow x < 12$

   Also $x > 0$. So the domain is $0 < x < 12$.

   Watch out

   A common mistake is to forget that $x$ cannot exceed half the smaller side. Here $x$ must be less than $12$, not $22.5$, because the width $24$ cm is the limiting factor.

3. Expand and simplify $V(x)$ for differentiation.

   First multiply the two binomials:

Then multiply by $x$:

4. Differentiate $V(x)$ with respect to $x$. $V'(x) = 1080 - 276x + 12x^2$ …

Method: Open-Box-From-a-Rectangular-Sheet Optimization (Two Unequal Sides)

This method extends the square-sheet box-folding technique to a rectangular sheet, where the length and width shrink by the same amount $2x$ but start from different original sizes — so the domain restriction becomes more subtle.

Steps

Step 1: Define the cut length and express both base dimensions

Let $x$ be the side of the square cut from each corner. For a sheet of length $l$ and width $w$ ($l > w$):

Step 2: Identify the domain using the SHORTER side, not the longer one

Both $l-2x>0$ and $w-2x>0$ must hold, giving $x < l/2$ and $x < w/2$. Since $w$ is the smaller dimension, $w/2$ is the tighter (binding) restriction — the true domain is $0 < x < w/2$, not the looser bound from the longer side.

Step 3: Write the volume function and expand for differentiation

$V(x) = x(l-2x)(w-2x)$ …

Common Mistakes

Mistake 1: Using the longer side to bound the domain instead of the shorter one

Why it's wrong: both sides impose a constraint ($x < l/2$ and $x < w/2$), but only the tighter one actually limits $x$ — using the looser bound from the longer side (e.g. $x<22.5$ instead of $x<12$) can let an invalid root slip through as if it were feasible. Correct approach: always take the smaller of the two half-side values as the true upper bound on $x$.

Mistake 2: Arithmetic error expanding $(l-2x)(w-2x)$

Why it's wrong: this expansion has four terms to track ($lw$, $-2lx$, $-2wx$, $4x^2$) and a sign or coefficient slip here propagates into a wrong quadratic when solving $V'(x)=0$. Correct approach: expand methodically and double-check by substituting a numeric value of $x$ into both the factored and expanded forms.

Mistake 3: Keeping the out-of-domain root as if it were a valid critical point …