Open-Top Box Design From Fixed Sheet Area

From Sheet Area to a Box Design

An open-top box uses one rectangular sheet for its bottom and four sides. With a fixed amount of sheet material, changing the dimensions changes the volume the box can hold. The design problem is to choose the dimensions that give the greatest volume without exceeding or wasting the available area.

Let \(x\) and \(y\) be the lengths of the bottom edges, and let \(h\) be the height. Assume the sheet has area \(S\), where \(S>0\). All three dimensions must be positive for the box to have positive volume. We will also consider zero dimensions when checking the feasible boundary.

The factors of \(2\) matter: an open-top box has no lid, but it still has two sides of each of the two side-panel sizes. The optimization is to maximize \(V\) subject to the area constraint and the feasibility conditions \(x,y,h\geq 0\).

Reduce the Constraint and Inspect Feasibility

For positive \(x\) and \(y\), the sheet-area equation determines the height:

Because \(x+y>0\), this height is positive exactly when \(xy<S\). Substituting it into the volume gives a two-variable expression:

This reduction is useful for understanding the feasible designs, but it does not by itself establish where the maximum occurs. In particular, the allowed positive values of \(x\) and \(y\) do not form a bounded region: one dimension can grow while the other shrinks. We must account for that possibility as well as the finite boundary cases.

Why a Global Maximum Exists

If any one of \(x,y,h\) is zero, then \(V=xyh=0\). There are feasible boxes with positive volume: for example, choose any \(x,y>0\) with \(xy<S\), and the formula above gives \(h>0\). Thus a maximizing design cannot lie on a finite boundary where one dimension is zero.

There is also a possible escape to unbounded dimensions. Set \(q=xy\). For every feasible positive box, \(0<q<S\) and \(x+y\geq 2\sqrt{q}\). Consequently,

The right-hand side tends to zero if \(q\) tends to \(0\) or to \(S\). Also, since \(q(S-q)\leq S^2/4\),

So if \(x+y\) grows without bound, the volume tends to zero. These estimates show that a sequence of feasible boxes whose volumes stay above a fixed positive value cannot approach \(q=0\), \(q=S\), or \(x+y=\infty\). Its values of \(x\) and \(y\) are bounded, and \(xy\) stays positive, so neither dimension can tend to zero. A convergent subsequence therefore has a positive feasible limit, with its height determined continuously by the area constraint. Hence the greatest volume is attained at a box with \(x,y,h>0\), not merely approached by a sequence of designs.

Find the Interior Candidate

At an interior constrained maximum, the Lagrange multiplier condition applies when the constraint gradient is nonzero. Here define

The constraint is \(g=S\). Its gradient is \(\nabla g=(y+2h,\ x+2h,\ 2x+2y)\), which is nonzero for positive dimensions. The necessary condition is therefore \(\nabla F=\lambda\nabla g\) for some real number \(\lambda\). Written component by component, it is

The first two equations force the two base edges to be equal. Subtracting the second equation from the first gives

Suppose instead that \(x\ne y\). Then \(h=\lambda\). Substituting this into the first Lagrange equation would give \(yh=h(y+2h)\), or \(0=2h^2\), which contradicts \(h>0\). Therefore \(x=y\).

Write \(x=y=a\). The third equation gives \(a^2=4\lambda a\), so \(\lambda=a/4\). The first equation then becomes

Using the area constraint, \(S=a^2+4ah\). Since \(h=a/2\), this reduces to \(S=3a^2\). The only positive candidate is \(a=\sqrt{S/3}\).

This is a global maximum, not just an interior candidate. The preceding feasibility argument shows that a maximum is attained at positive dimensions. Every such maximum must satisfy the Lagrange equations, which force the dimensions just found. The candidate is feasible and has positive volume, while every finite boundary box has zero volume. Thus no boundary design or unbounded sequence can produce a greater value.

Worked Example: A Sheet with Area 300 Square Centimetres

Worked Example: A Sheet with Area 192 Square Metres

Worked Example: Check an Unequal-Base Design

A Common Area-Counting Mistake

A frequent error is to include only one side of each size, writing the constraint as \(xy+xh+yh=S\). That omits two of the four side panels and describes a different material-use model. For example, with \(S=300\text{ cm}^2\), the incorrect constraint would give a square-base candidate \(a=10\text{ cm}\), \(h=10\text{ cm}\): it predicts volume \(1000\text{ cm}^3\). But in the actual open-top box, that design uses

not \(300\text{ cm}^2\). The correction is to count both panels of each side size. With the correct constraint, the \(300\text{ cm}^2\) optimum is \(10\text{ cm}\) by \(10\text{ cm}\) by \(5\text{ cm}\), and its area check is \(100+200=300\text{ cm}^2\).

Why the Design Result Is Useful

The result gives a practical rule for converting sheet area into an optimal box: make the base square and choose a height equal to half its side length. The square base is not an assumption built into the model; it follows from the optimization. The proof also explains why merely finding an interior critical point is not enough in a design problem. Feasibility includes degenerate edges, and a dimension can become arbitrarily large unless escaping designs are addressed. Here the boundary and escape checks establish that the candidate really is the global best design.