From a Flat Sheet to a Box
A rectangular sheet can be turned into an open-top box by cutting an equal square from each corner and folding up the sides. Choosing a larger square makes the box taller, but it also makes its base smaller. The greatest volume balances these competing effects.
To find that volume, first express all three box dimensions using one variable: the side length of each corner square. Then identify which cut sizes are physically possible, find the interior critical points, and compare any valid candidates with the boundary cases.
Deriving the Volume Function and Its Domain
Each corner cut removes a strip of width \(x\) from both ends of each side of the sheet. Thus the base loses \(x\) at one end and \(x\) at the other, making its dimensions \(L-2x\) and \(W-2x\). After folding, the height is \(x\), so multiplying the three dimensions gives the volume.
A feasible cut must be positive, and neither base dimension can be negative. Therefore \(x>0\), \(L-2x\geq 0\), and \(W-2x\geq 0\). In particular, if \(m=\min(L,W)\), the feasible interval, including degenerate endpoint cases, is \(0\leq x\leq m/2\). At either endpoint the volume is zero: at \(x=0\), the box has no height; at \(x=m/2\), at least one base dimension is zero. A nondegenerate box requires \(0<x<m/2\).
Why the Critical-Point Test Finds the Maximum
The volume function is a polynomial, so it is continuous on the closed, bounded feasible interval \([0,m/2]\). By the Extreme Value Theorem, it attains an absolute maximum there. Since both boundary volumes are zero, while every feasible interior cut produces positive volume, the absolute maximum must occur at an interior point.
At an interior maximum of a differentiable function, the derivative is zero. This gives a way to locate possible maximum-volume cut sizes; the domain and boundary values are still needed to decide which candidates are physically relevant and whether they give the absolute maximum.
Solving \(V'(x)=0\) gives the two algebraic critical-point candidates
Not every root is necessarily feasible. For example, a root greater than \(m/2\) would make at least one base dimension negative, so it cannot represent a box. The calculation must be followed by a check against the feasible interval.
Worked Example: A 20-by-14 Sheet
Worked Example: A 20-by-14 Sheet
A sheet measures \(20\) units by \(14\) units. Find the cut size that gives the greatest box volume, and find that volume.
Let \(x\) be the cut size. The base dimensions are \(20-2x\) and \(14-2x\), and the height is \(x\). The shorter sheet dimension limits the cut, so \(0\leq x\leq 7\). The volume function is
Differentiate and set the derivative equal to zero:
The quadratic equation has roots
Using \(\sqrt{79}\approx 8.888194\), the roots are approximately \(2.704\) and \(8.629\). Only \(2.704\) lies in the feasible interval \([0,7]\). The other root would make the \(14-2x\) side of the base negative.
The boundary volumes are \(V(0)=0\) and \(V(7)=7(6)(0)=0\). At the feasible critical point, using \(\sqrt{79}\approx8.888194\) and rounding dimensions to six decimal places, the height is \(x\approx2.703935\), and the base dimensions are
Thus \(V(x)\approx(2.703935)(14.592130)(8.592130)\approx339.013\) cubic units, rounded to three decimal places. The critical-point calculation is checked by substitution: the chosen value is \(x=(17-\sqrt{79})/3\), which satisfies \(3x^2-34x+70=0\), so \(V'(x)=4(3x^2-34x+70)=0\). Its positive volume exceeds both boundary values, so it gives the absolute maximum.
Worked Example: A Square Sheet
Worked Example: An 18-by-18 Sheet
An \(18\)-unit square sheet is used to make an open-top box. Find the maximum volume and explain how the boundary critical point is treated.
The cut size must satisfy \(0\leq x\leq9\). Both base dimensions are \(18-2x\), so
Differentiate and factor:
The critical points are \(x=3\) and \(x=9\). The first is interior; the second is the right endpoint of the feasible interval and gives a collapsed base. At the interior point, the height is \(3\) units and both base dimensions are \(18-2(3)=12\) units. The volume is \(3(12)(12)=432\) cubic units. At the boundaries, \(V(0)=0\) and \(V(9)=9(0)(0)=0\).
For a derivative check, \(V'(3)=12(3-3)(3-9)=0\). The volume \(432\) is greater than either boundary value, so the maximum is \(432\) cubic units, attained with a \(3\)-unit cut. The root \(x=9\) is not a second maximum-volume box: it is a degenerate boundary case with zero volume.
Worked Example: Rejecting an Infeasible Critical Point
Worked Example: A 10-by-6 Sheet
A \(10\)-unit-by-\(6\)-unit sheet is cut to make a box. Find the feasible critical point and check the boundary cases.
Since the shorter sheet side is \(6\), the feasible interval is \(0\leq x\leq3\). The volume function and its derivative are
Setting the derivative equal to zero gives
Since \(\sqrt{19}\approx4.358899\), these values are approximately \(1.214\) and \(4.120\). Only \(1.214\) is feasible; the larger value exceeds \(3\). At the feasible root, the height is approximately \(1.213700\), and the base dimensions are \(10-2x\approx7.572599\) and \(6-2x\approx3.572599\). Their product with the height gives \(V(x)\approx32.835\) cubic units, rounded to three decimal places.
The derivative check follows from the root equation: \(3x^2-16x+15=0\) at \(x=(8-\sqrt{19})/3\), so \(V'(x)=0\). At the feasible boundaries, \(V(0)=0\) and \(V(3)=3(4)(0)=0\). The positive interior volume therefore gives the absolute maximum. The other algebraic root does not describe a box.
Common Pitfall: Using the Wrong Feasible Bound
A cut size must satisfy both \(x\leq L/2\) and \(x\leq W/2\). Checking only the longer dimension can produce a seemingly valid algebraic answer that is physically impossible.
For the \(20\)-by-\(14\) sheet, using only \(x\leq20/2=10\) might lead someone to accept the larger critical point \(x\approx8.629\). But then the purported base side \(14-2x\) is approximately \(14-17.258=-3.258\) units. A negative side length cannot describe a box. The correct limit is \(x\leq\min(20,14)/2=7\), which excludes this root.
A related modeling error is to write the base dimensions as \(L-x\) and \(W-x\). There are two corner cuts along each base direction, one at each end, so each dimension decreases by \(2x\), not \(x\). A reliable model checks the folded geometry before optimizing the resulting formula.
A Reliable Optimization Process
The calculation combines geometry, algebra, and the derivative test. Keeping those tasks in order helps prevent an algebraic root from being mistaken for a physically possible answer.
Let \(x\) be the corner-square side length. Write the height as \(x\) and the base dimensions as \(L-2x\) and \(W-2x\).
Use both sheet dimensions to obtain \(0\leq x\leq\min(L,W)/2\). Interior, nondegenerate boxes have strict inequalities.
Find \(V'(x)\), set it equal to zero, and solve for all critical-point candidates.
Discard roots outside the interval, evaluate the volume at feasible candidates, and check the endpoint values.
State the cut size and maximum volume with units. Confirm that the dimensions are nonnegative and that the selected interior value satisfies \(V'(x)=0\).
Check Your Understanding
For each question, keep the physical restrictions on the cut size in view as you calculate.
- For a sheet measuring \(16\) units by \(10\) units, write the volume function and its feasible interval.
- Why does cutting a square of side length \(x\) reduce each base dimension by \(2x\), rather than \(x\)?
- For an \(18\)-by-\(18\) sheet, why does the critical point \(x=9\) not produce a positive-volume box?
- For a rectangular sheet, why must any positive-volume absolute maximum occur at an interior critical point?
- What physical problem arises if a critical-point candidate is larger than \(\min(L,W)/2\)?
