Q.A square piece of tin of side $18$ cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible.

Concept: Optimization Word Problem — maximizing volume by choosing the cut size.

Let the side of the square cut from each corner be $x$ cm. After cutting and folding, the base of the box becomes a square of side $18 - 2x$, and the height is $x$. The volume is:

We need $0 < x < 9$ for a valid box. Differentiate and set $V'(x) = 0$: …

We start with a flat square of tin, 18 cm on each side. The goal is to cut identical squares from the four corners, then fold up the flaps to form an open-top box. The question asks: what size square should we cut so that the box holds the greatest volume?

This is a classic optimization word problem. The key idea is to express the volume of the box as a function of the cut size, then use differentiation to find where that function reaches its maximum. Because the box has no top, the height of the box equals the side of the cut-out square. The length and width of the base shrink by twice the cut size (since we remove a square from both ends of each side).

Let’s work through it step by step.

1. Define the variable.

   Let $x$ cm be the side length of the square cut from each corner. Since the original side is 18 cm, after cutting, the base of the box becomes a square of side $18 - 2x$ cm. The height of the box is exactly $x$ cm (the flaps are folded up).

2. Write the volume function.

   Volume of a box = (base area) × (height).

   Base area = $(18 - 2x)^2$, height = $x$.

   So

The domain: $x$ must be positive (we cut something) and less than 9 (otherwise the base vanishes or becomes negative). So $0 < x < 9$.

3. Differentiate to find critical points. Expand first (or use product rule). Let’s expand:

Differentiate:

Set $V'(x) = 0$:

Divide through by 12:

Factor:

So $x = 3$ or $x = 9$.

$x = 9$ is at the boundary of the domain (base becomes zero), so it gives zero volume — not a maximum. The only interior critical point is $x = 3$.

4. Confirm it’s a maximum. Use the second derivative test: $V''(x) = 24x - 144.$ …

Method: Open-Box-From-a-Sheet Optimization (Cut-and-Fold Volume Problems)

This method solves the classic "cut squares from the corners of a sheet, fold up the flaps, maximize the box's volume" family of problems.

Steps

Step 1: Define the cut length as the single variable

Let $x$ be the side length of the square cut from each corner. This single variable determines every dimension of the resulting box.

Step 2: Express the box's length, width, and height in terms of $x$

Because a square of side $x$ is removed from both ends of each side of the sheet, each side of the base shrinks by $2x$ (not $x$), while the height of the box equals $x$ (the folded-up flap). For a square sheet of side $a$:

Step 3: Write the volume as a function of $x$ and identify its valid domain

The domain is restricted to $0 < x < a/2$, since the base side must stay positive. …

Common Mistakes

Mistake 1: Writing the base side as $a - x$ instead of $a - 2x$

Why it's wrong: a square is cut from each of the four corners, which removes a strip of width $x$ from both ends of every side — so each side shrinks by $2x$ total, not $x$. Using $a-x$ silently changes the whole volume function and gives a wrong critical point. Correct approach: always sketch the folded sheet and note that two corner-cuts affect each side.

Mistake 2: Forgetting the domain restriction and accepting the boundary root as the answer

Why it's wrong: the derivative's factored form typically produces two roots, one of which sits exactly at the edge of the valid domain (where the base side becomes zero, giving zero volume) — that root can never be a genuine maximum, since it makes the box degenerate. Correct approach: always check $0 < x < a/2$ and discard any root outside (or at the edge of) that range before comparing. …