Maximizing The Volume Of An Open Box

From a Flat Sheet to an Open Box

Cutting equal squares from the corners of a rectangular sheet and folding up the remaining sides is a standard way to make an open-top box. The optimization question is to find the cut size that gives the largest volume. The dimensions of the sheet constrain the cut, and the resulting volume depends on all three dimensions of the folded box.

Let the sheet have length \(L\) and width \(W\), with \(L\ge W>0\). Cut a square of side \(x\) from each of its four corners. After folding, the height of the box is \(x\). Each horizontal dimension is reduced at both ends, so the base has length \(L-2x\) and width \(W-2x\).

Here \(L\), \(W\), and \(x\) are lengths in the same units. The volume is measured in cubic units. The factor of \(2\) in each base dimension matters: a cut removes material from both ends of that dimension.

Feasible Cut Sizes

A physical box with positive height and positive base dimensions requires \(x>0\), \(L-2x>0\), and \(W-2x>0\). Since \(L\ge W\), the width constraint is the more restrictive one. Thus the cut sizes that produce a nondegenerate box satisfy

For comparing boundary values, it is useful to extend the volume formula to the closed interval \(0\le x\le W/2\). At \(x=0\), the height is zero, so the volume is zero. At \(x=W/2\), the base width is zero, so the volume is again zero. Neither endpoint produces a positive-volume box, but both are legitimate boundary values of the extended function.

Substituting the dimensions into the volume formula gives the objective function

This is a continuous function on the closed feasible interval. Its maximum on that interval must occur at a boundary point or at an interior critical point, where the derivative is zero. The boundary values alone cannot be the maximum if an interior cut gives positive volume.

Finding the Critical Points

Differentiate the expanded cubic:

Setting the derivative equal to zero gives a quadratic equation. The quadratic formula produces two roots. Define \(D=\sqrt{L^2-LW+W^2}\), which is positive because \(L\) and \(W\) are positive. The roots are

Having two algebraic roots does not mean both are feasible. Only a root in the interval \(0<x<W/2\) can describe a nondegenerate box. The shorter sheet dimension determines which root is relevant.

Why the Feasible Critical Point Is the Global Maximum

We can establish the theorem and the maximum claim directly. First, \(D<L+W\): both quantities are positive, and

Thus \(x_->0\). To show \(x_-<W/2\), rearrange the inequality to \(L-2W<D\). If \(L-2W\le0\), this holds because \(D>0\). If \(L-2W>0\), then

In this case \(L>2W\), so \(L>W\) and the right-hand side is strictly positive. Both quantities being compared are positive, hence \(D>L-2W\). This proves \(x_-<W/2\).

For the other root, \(x_+\ge W/2\) is equivalent to \(D\ge2W-L\). If \(2W-L\le0\), that is immediate. If \(2W-L>0\), then

Since both sides of \(D\ge2W-L\) are then nonnegative, the squared inequality implies the desired inequality. This proves that \(x_+\) is not an interior feasible critical point. When \(L=W\), it equals the boundary value \(W/2\); when \(L>W\), it lies beyond that boundary.

Finally, the derivative factors using its two roots:

For \(0\le x<x_-\), both factors in parentheses are negative, so \(V'(x)>0\). For \(x_-<x<W/2\), the first factor is positive and the second negative, so \(V'(x)<0\). Thus the volume increases up to \(x_-\) and decreases after it across the feasible interval. The endpoint volumes are both zero, while \(V(x_-)>0\) because all three dimensions are positive there. Consequently, \(x_-\) gives the global maximum, not merely a local maximum.

Worked Example: A 16-by-10 Sheet

Worked Example: A Square Sheet

Worked Example: Rejecting an Infeasible Root

A Common Modeling Mistake

A frequent error is to subtract \(x\), rather than \(2x\), from each base dimension. For the \(16\text{ cm}\) by \(10\text{ cm}\) sheet, that mistake would assign a cut of \(2\text{ cm}\) the dimensions \(14\text{ cm}\) and \(8\text{ cm}\), and give the incorrect volume

But each horizontal dimension loses a corner square at both ends. The correct dimensions are \(16-2(2)=12\) cm and \(10-2(2)=6\) cm, giving \(2\cdot12\cdot6=144\text{ cm}^3\). The correction is to track the two removed lengths separately in each direction before writing the volume function.

A second important check is to test every critical point against the feasible interval. A derivative equation describes stationary points of the algebraic function, but a root outside the interval cannot describe a possible box. The \(8\)-by-\(5\) example shows why finding the roots is not the end of the optimization problem.

Why Boundary Values Matter

The maximum-volume cut is an interior point, but checking the boundaries completes the global comparison. The physical domain excludes both endpoints because each gives a degenerate shape rather than a box of positive height and base area. Nevertheless, the volume formula extends continuously to the closed interval, and its values at both endpoints are zero. The interior critical point has positive volume, so it exceeds both boundary values.

This comparison also distinguishes the problem’s actual geometry from the closed-interval method used to establish a maximum. The extended interval is a tool for checking all possible limiting cases; it does not make a zero-height or zero-width object a valid box. The final maximizing cut lies strictly inside the physical domain, so it is attainable.