13. The figure is of a square sheet measuring 10cm x 10cm. Squares with side a cm are
cut out from the four corners. The resulting sheet is bent into a box. Find the value of
a that maximizes the volume of the box so formed.
a. 5
b. 3/5
c. 5/3
d. 4

Introduction

This article explains how to determine the maximum volume of a box from a 10 cm square sheet by cutting out equal squares of side a from each corner and folding up the sides.
Using calculus (maxima and minima), the volume is written as a function of a, differentiated, and optimized within the allowed range of a.

Problem statement and model

A square sheet of side 10 cm is used. Squares of side a cm are cut from all four corners, and the remaining flaps are folded up to form an open box.

• Length of box after folding: 10 − 2a cm (because a is removed from both left and right).
• Width of box after folding: 10 − 2a cm (same reasoning).
• Height of box: a cm (the side length of the cut square).

So the volume as a function of a is

V(a) = a(10 − 2a)².

The physical constraints are:

• a > 0 (positive height).
• 10 − 2a > 0 ⇒ a < 5 (base dimensions must remain positive).

Thus the domain of a is 0 < a < 5.

Calculus: finding the value of a

First expand the volume function:

(10 − 2a)² = 100 − 40a + 4a²,

so

V(a) = a(100 − 40a + 4a²) = 4a³ − 40a² + 100a.

Differentiate with respect to a:

V′(a) = 12a² − 80a + 100.

Set derivative equal to zero to get critical points:

12a² − 80a + 100 = 0.

Divide throughout by 4:

3a² − 20a + 25 = 0.

Solve this quadratic using the quadratic formula:

a = [20 ± √(20² − 4 · 3 · 25)] / (2 · 3)
= [20 ± √(400 − 300)] / 6
= (20 ± 10) / 6.

Thus:

• a = 30 / 6 = 5 cm,
• a = 10 / 6 = 5 / 3 cm.

Apply the domain restriction 0 < a < 5.

• At a = 5 cm, base side 10 − 2a = 0, so the “box” collapses and the volume is zero, which cannot be a maximum.
• At a = 5 / 3 cm, the dimensions are all positive and the box exists, so this is the feasible critical point.

A second‑derivative test, or comparison of volumes near this value, confirms that a = 5 / 3 cm gives the maximum volume.

Therefore, the value of a that maximizes the volume is
a = 5 / 3 cm.

Checking each multiple‑choice option

The given options are:

• A) 5
• B) 3/5
• C) 5/3
• D) 4

Evaluate why each is right or wrong within the domain 0 < a < 5.

Option A: a = 5

If a = 5, then each base dimension is 10 − 2a = 0, so the volume
V = 5 · 0² = 0.
The box degenerates to a line, so this gives minimum, not maximum, volume.
Hence option A is incorrect.

Option B: a = 3/5

If a = 3/5 ≈ 0.6, then base side 10 − 2a ≈ 8.8 cm and volume is positive but less than the maximum volume at a = 5 / 3.
This value does not satisfy the derivative‑zero condition for maxima, so option B is incorrect.

Option C: a = 5/3

For a = 5 / 3 ≈ 1.67, the derivative V′(a) is zero, the second derivative is negative, and the box has dimensions (10 − 10/3) × (10 − 10/3) × 5/3, all positive.
This choice yields the unique interior maximum on 0 < a < 5, so option C is correct.

Option D: a = 4

If a = 4, each base side becomes 10 − 8 = 2 cm and the height is 4 cm, so the volume is 4 · 2 · 2 = 16 cm³, which is smaller than the maximum at a = 5 / 3.
Also V′(4) ≠ 0, so option D is incorrect.

Final answer

The value of a that maximizes the volume of the box formed from a 10 cm × 10 cm square sheet by cutting equal squares from the four corners and folding up the sides is
a = 5 / 3 cm, corresponding to option C) 5/3.