How to minimize the cost of an open-top box with a square base

Expert intro: This is a classic constrained optimization problem: the volume is fixed, but the material costs are not the same for the base and the sides. That means the cheapest box is not the one with the most symmetric dimensions, but the one that balances surface area against the different unit prices.

Detailed walkthrough

Let the square base have side length $x$ cm and height $h$ cm.

1) Use the volume constraint

The volume is fixed at 500 cm$^3$, so

$x^2h=500$

which gives

$h=\frac{500}{x^2}.$

2) Write the cost function

• Base area: $x^2$, at $0.10$ dollars/cm$^2$ gives cost $0.10x^2$
• Four sides: each side has area $xh$, so total side area is $4xh$. At $0.05$ dollars/cm$^2$, side cost is

$0.05(4xh)=0.20xh.$

So the total cost is

$C(x)=0.10x^2+0.20x\left(\frac{500}{x^2}\right)=0.10x^2+\frac{100}{x}.$

3) Differentiate and find critical points

$C'(x)=0.20x-\frac{100}{x^2}.$

Set $C'(x)=0$:

$0.20x-\frac{100}{x^2}=0$

$0.20x^3=100$

$x^3=500$

$x=\sqrt[3]{500}.$

This is the only critical point for $x>0$.

4) Confirm it is a minimum

Differentiate again:

$C''(x)=0.20+\frac{200}{x^3}.$

For $x>0$, $C''(x)>0$, so the critical point gives a minimum cost.

5) Find the height

$h=\frac{500}{x^2}=\frac{500}{(\sqrt[3]{500})^2}=\sqrt[3]{500}.$

So the minimum-cost box is actually a cube.

Answer

• Base side length: $x=\sqrt[3]{500}\approx 7.94$ cm
• Height: $h=\sqrt[3]{500}\approx 7.94$ cm

The critical point occurs at $x=\sqrt[3]{500}$, and it gives the minimum cost.

💡 Pitfall guide

A common mistake is forgetting that the four side faces together have area $4xh$, not $xh$. Another easy slip is mixing up the material cost rates: the base is more expensive per square centimetre than the sides, so the base term must be $0.10x^2$, not $0.05x^2$. Also, when checking critical points, remember the domain is $x>0$ because the box must have a positive base length.

🔄 Real-world variant

If the cost rates changed, the minimizing box would usually no longer be a cube. In general, with base cost $c_b$ and side cost $c_s$, the cost function becomes

$C(x)=c_bx^2+\frac{200c_s}{x}$

after using $h=500/x^2$. Different prices shift the balance between a wide base and a tall box. If the volume were larger, the optimal dimensions would scale up, but the same optimization method still works.

🔍 Related terms

constrained optimization, cost function, critical point