Cutting a can from a metal sheet to maximize volume

Question:

Consider the following square sheet of metal (grey) with the side s = 10 cm. From it, we want to cut the parts of a cylindrical can with a lid and a bottom (bblack). Determine the radius and height of the can when the volume is maximal.

The rectangle and circles have to be cut in entire, singular pieces, so no trickery to minimize waste and no diagonal placement of the rectangle.

Solution has to use optimization / differentiation.

Attempted solution:

In order for this to work out the circumference of the circles and one side of the rectangle has to be identical. Otherwise they will not fit together in a cylindrical can.

This leads us to two cases that needs to be treated separately.

Case 1: The rectangle is standing upright.

Case 2: The rectangle is lying on its longest side.

There are two cases that needs to be considered separately.

Case 1:

The right-hand side of the above image must be the circumference of a circle, so

$$s_1 = 2r\pi$$

The bottom-side must be the height of the cylinder plus two radii of the circles:

$$s_1 = h + 2r$$

Putting these together becomes:

$$2r\pi = h + 2r = 10$$

Solving for h:

$$h = 10 - 2r$$

The volume of the can:

$$V = r^2\pi h$$

Adding in the earlier formula to get rid of h:

$$V = r^2\pi h = r^2\pi (10 - 2r) = 10r^2\pi - 2r^3\pi$$

Taking the derivative:

$$V' = 20\pi r - 6\pi r^2$$

Setting derivative equal to zero:

$$V' = 0 \Rightarrow r = \frac{10}{3}$$

(neglecting trivial r = 0 solution)

Height is then:

$$h = 10 - 2r = 10 - \frac{20}{3} = \frac{10}{3}$$

But this is not really possible given the side of the metal sheet cannot be more than 10 cm and $$2r\pi > 10$$.

Case 2:

The bottom becomes:

$$2r + 2r\pi = 10 \Rightarrow r= \frac{5}{1+\pi}$$

The right side becomes:

$$h = 10$$

However, none of these answers match the solution:

$$r = \frac{5}{\pi}$$

$$h = 10(1-\frac{1}{\pi})$$

Did I get the starting point for the possible cases right? Where did I go wrong? How come it appears that no calculus is needed in the second step?

• No need to consider the "two cases" (The rectangle is standing upright vs lying on its longest side) because the sheet is square. – leonbloy Jul 25 at 17:15
• @leonbloy are there are more than one potential case to consider? – MathInferno Jul 25 at 17:19
• @leonbloy: mathinferno is distinguishing the two cases where the vertical edge of the black rectangle is the height of the cylinder or its corcumference. – TonyK Jul 29 at 12:20

Yes, there are two cases. In your case 1 the side of the square equals the circumference of the lid; in case 2 the side of the square equals the height of the can.

In case 2 the volume of the can is a function of $$r^2$$ alone, so volume is maximized when $$r$$ is as large as possible. This is the reason why no calculus is required. The value of $$r$$ in this case is constrained by $$2r + 2\pi r\le 10$$, as you've shown.

The other answers have addressed case 1. It remains to check that the max volume under case 2 is less than the max volume under case 1.

Since the rectangle must be cut parallel to the sides of the square, the circumference of the circle is limited to the side of the square. So $$2\pi r \leq 10$$ or $$r \leq 5/\pi$$.

At this limit, the height is $$h = 10 - 10/\pi.$$

The volume $$V=\pi r^2h$$. Since any increase in height equals any decrease in radius, the limiting values provide the maximum volume because the volume is quadratic in the radius, but only linear in the height.

• It is a nice, clean solution. Thanks! For clarification, I have added that the question should be solved with differentiation / optimization. – MathInferno Jul 25 at 17:13

Calling $$b, s$$ the metal sheet dimensions, $$r$$ the tip radius and assuming an arrangement as depicted in the OP figure we have

$$V = \pi r^2 s$$ with the associated lagrangian $$L(r,\lambda,\epsilon) = V + \lambda_1(4r-s+\epsilon_1^2)+\lambda_2(2r+2\pi r -b+\epsilon_2^2)$$

Here $$\epsilon_i$$ are slack variables to transform the inequalities into equations.

Now solving for the stationary points we have

$$\nabla L = 0 = \left\{ \begin{array}{rcl} 2 \pi r s+4 \lambda _1+2(1+\pi ) \lambda _2&=&0 \\ \epsilon_1^2+4 r-s&=&0 \\ \epsilon_2^2-b+2 \pi r+2 r&=&0 \\ \epsilon_1 \lambda _1&=&0 \\ \epsilon_2 \lambda _2&=&0 \\ \end{array} \right.$$

and after solving we get

$$\left[ \begin{array}{c} V&r&\lambda_1&\lambda_2&\epsilon_1^2&\epsilon_2^2 \\ \frac{b^2 \pi s}{4 (1+\pi )^2}&\frac{b}{2(1+ \pi) }&0&-\frac{b \pi s}{2 (1+\pi )^2}&s-\frac{2 b}{1+\pi}&0 \\ \frac{\pi s^3}{16}&\frac{s}{4}&-\frac{\pi s^2}{8}&0&0&b-\frac{1}{2} (1+\pi ) s \\ \end{array} \right]$$

NOTE

$$\epsilon_i = 0$$ means that the $$i-$$constraint is verified at the boundary