Optimizing cylinder volume from a rectangular sheet metal

Expert intro: This is an optimization problem with a volume function and a perimeter constraint. The key is to express everything in one variable, then use calculus to locate the maximum.

Detailed walkthrough

Set up the variables

Let the rectangle have side lengths $x$ and $y$, where the sheet is rolled to form an open cylinder. From the problem setup,

$x=2\pi r, \qquad y=h,$

and the perimeter constraint is

$2x+2y=50.$

Substitute $x=2\pi r$ and $y=h$:

$2(2\pi r)+2h=50,$

which simplifies to

$4\pi r+2h=50.$

So

$h=25-2\pi r.$

Write the volume as one variable

The volume of a cylinder is

$V=\pi r^2h.$

Substitute the expression for $h$:

$V(r)=\pi r^2(25-2\pi r).$

Expanding gives

$V(r)=25\pi r^2-2\pi^2r^3.$

Differentiate and find the critical point

Differentiate with respect to $r$:

$V'(r)=50\pi r-6\pi^2r^2.$

Factor:

$V'(r)=2\pi r(25-3\pi r).$

Set the derivative equal to zero:

$2\pi r(25-3\pi r)=0.$

The nonzero critical point is

$r=\frac{25}{3\pi}.$

Then

$h=25-2\pi\left(\frac{25}{3\pi}\right)=25-\frac{50}{3}=\frac{25}{3}.$

Compute the maximum dimensions

So the rectangle dimensions are

$x=2\pi r=2\pi\left(\frac{25}{3\pi}\right)=\frac{50}{3}\text{ cm},$

and

$y=h=\frac{25}{3}\text{ cm}.$

Numerically,

$x\approx 16.67\text{ cm}, \qquad y\approx 8.33\text{ cm}.$

The maximum volume is

$V=\pi\left(\frac{25}{3\pi}\right)^2\left(\frac{25}{3}\right)=\frac{15625}{27\pi}\approx 184.1\text{ cm}^3.$

Final answer

The rectangle that maximizes the cylinder volume has dimensions

$\boxed{\frac{50}{3}\text{ cm} \times \frac{25}{3}\text{ cm}}$

and the maximum volume is

$\boxed{\frac{15625}{27\pi}\text{ cm}^3\approx 184.1\text{ cm}^3}.$

💡 Pitfall guide

A common mistake is to use the perimeter equation as $x+2y=50$, which would be incorrect because a rectangle has perimeter $2x+2y$. Another issue is mixing up the rolled dimension: the side that becomes the circumference must be written as $2\pi r$, not $\pi r^2$ or another length. Some students also stop after finding a critical point without checking whether it is a maximum. Here the second derivative or the shape of the volume function confirms that the critical point gives the greatest volume. Finally, keep the dimensions consistent in centimeters throughout the calculation.

🔄 Real-world variant

If the rectangle were rolled the other way, so that the longer side became the height instead of the circumference, the constraint would still be the same but the variable assignment would change. For example, if $x=h$ and $y=2\pi r$, then the algebra would look different, yet the optimization idea would remain identical. If the perimeter were 60 cm instead of 50 cm, the same derivative method would work, but the optimal dimensions would scale to the new constraint. The key is that one side always becomes the circumference and the other becomes the cylinder height.

🔍 Related terms

cylinder volume, derivative test, constraint optimization