Minimizing the cost of an open-top rectangular box

Key concept: This is a classic constrained optimization problem: one variable determines the box dimensions, and the cost function is minimized with calculus.

Step by step

Set up the constraint and cost function

Let the square base have side length $x$ and height $h$. Since the volume is fixed at 216 cubic inches,

$x^2h=216,$

so

$h=\frac{216}{x^2}.$

The cost is different for the base and the sides. The base costs 20 cents per square inch, so its cost is

$0.2x^2.$

The sides have total area $4xh$, and they cost 30 cents per square inch, so the side cost is

$0.3(4xh)=1.2xh.$

That gives the cost function

$C(x)=0.2x^2+1.2x\left(\frac{216}{x^2}\right)=0.2x^2+\frac{259.2}{x}.$

Differentiate and find the critical point

Differentiate with respect to $x$:

$C'(x)=0.4x-\frac{259.2}{x^2}.$

Set the derivative equal to zero:

$0.4x-\frac{259.2}{x^2}=0.$

Multiply by $x^2$:

$0.4x^3=259.2.$

So

$x^3=648,$

and therefore

$x=\sqrt[3]{648}=8.64.$

Since the cost function has a positive second derivative for $x>0$, this critical point gives a minimum.

Compute the height and interpret the result

Now substitute $x=8.64$ into the volume equation:

$h=\frac{216}{8.64^2}\approx 2.89.$

So the dimensions that minimize cost are

$8.64\text{ in} \times 8.64\text{ in} \times 2.89\text{ in}.$

This makes sense physically: when side material is more expensive than base material, the optimal box is relatively wider and shorter rather than tall and narrow.

Common calculus takeaway

The key idea is to rewrite the cost in one variable before differentiating. If you leave both $x$ and $h$ in the expression, you cannot minimize efficiently. The volume equation is what lets you eliminate $h$ and build a single-variable optimization problem.

Pitfall alert

A common mistake is treating the side area as $4x^2$ instead of $4xh$. The sides are rectangles, not squares, so each side has area $xh$, and there are four of them. Another frequent error is forgetting to convert cents into dollars consistently. The units do not change the minimizing dimensions, but they do matter if you want a correct cost function. Finally, some students stop after finding a critical point without checking whether it is a minimum; here, the second derivative test or the shape of the cost graph confirms the result.

Try different conditions

If the box were not square-based and instead had a rectangular base with length $x$ and width $y$, the setup would change to $xyh=216$ and the cost would depend on both $x$ and $y$. For example, if the problem became: "An open-top box has base dimensions $x$ by $2x$ with volume 216," then the single-variable cost function would be different, and the minimizing dimensions would not be equal on the base. The optimization method is the same, but the algebra changes because the geometry of the base changes.

Further reading

constrained optimization, cost function, open-top box