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Optimization

Calculus Optimization Problem:
What dimensions minimize the cost of an open-topped can?

An open-topped cylindrical can must contain V cm$^3$ of liquid. (A typical can of soda, for example, has V = 355 cm$^3$.) What dimensions will minimize the cost of metal to construct the can?


Calculus Solution

We’ll use our standard Optimization Problem Solving Strategy to develop our solution. (Link will open in a new tab.)

Stage I: Develop the function.

Your first job is to develop a function that represents the quantity you want to optimize. It can depend on only one variable. The steps:
A cylindrical open-topped can has height h and radius r, and is open at the top
1. Draw a picture of the physical situation.

See the figure. We’ve called the radius of the cylinder r, and its height h.


2. Write an equation that relates the quantity you want to optimize in terms of the relevant variables.

We want to minimize the amount of metal we use, which is to say we want to minimize the area of the can. The can consists of a cylinder of surface area $A_\text{cylinder} = (2\pi r)(h)$, and a bottom piece that has area $A_\text{bottom} = \pi r^2$. The can has no top. Its total area is thus:
\begin{align*}
A &= A_\text{cylinder} + A_\text{bottom}\\[4px] &= 2\pi rh + \pi r^2
\end{align*}

3. If necessary, use other given information to rewrite your equation in terms of a single variable.

The can’s area A currently depends on two variables, r and h. In order to proceed, we must use other information we’re given to rewrite the area in terms of just one of those variables. Let’s choose r as that single variable.

We must then eliminate the height h as a variable. To do so, we note that the can must hold volume V of liquid. Since the volume of a cylinder of radius r and height h is
$$V = \pi r^2 h$$

We can therefore solve for h:
$$h = \dfrac{V}{\pi r^2}$$

Substituting this expression for h into the expression for the can’s surface area:
\begin{align*}
A &= 2\pi rh + \pi r^2 \\[8px] &= 2\pi r\left( \dfrac{V}{\pi r^2} \right) + \pi r^2 \\[8px] &= 2\cancel{\pi} \cancel{r}\left( \dfrac{V}{\cancel{\pi} r\cancel{^2}} \right) + \pi r^2 \\[8px] &= \frac{2V}{r} + \pi r^2
\end{align*}
The expression for A is now a function of the single variable r. (Remember that V is a constant that just tells us how much liquid the can must hold.)

graph of the can's area as a function of its radius
We’ve graphed the function (with $V = 355 \text{ cm}^3$), a step you probably wouldn’t do yourself — but we want to emphasize that everything you’ve done so far is to create a function that you’re now going to minimize. One choice would be to closely examine the graph to determine the value of r that minimizes A . . . but instead we’re going to use the max/min techniques you learned recently!

Stage II: Maximize or minimize the function.

You now have a standard max/min problem to solve.
4. Take the derivative of your equation with respect to your single variable. Then find the critical points.
\begin{align*}
\frac{dA}{dr} &= \frac{d}{dr} \left[ \frac{2V}{r} + \pi r^2 \right] \\ \\
&= -\frac{2V}{r^2} + 2\pi r
\end{align*}

The critical points occur when $\dfrac{dA}{dr} = 0$:
\begin{align*}
\frac{dA}{dr} = 0 &= -\frac{2V}{r^2} + 2\pi r \\ \\
\frac{2V}{r^2} &= 2\pi r \\
r^3 &= \frac{V}{\pi} \\ \\
r &= \sqrt[3]{\frac{V}{\pi}}
\end{align*}

5. Justify your maxima or minima either by reasoning about the physical situation, or with the first derivative test, or with the second derivative test.

Let’s examine the second derivative. Above we found the first derivative:

$$\frac{dA}{dr} = -\frac{2V}{r^2} + 2\pi r $$
So
\begin{align*}
\dfrac{d^2A}{dr^2} &= \dfrac{d}{dr}\left( -\frac{2V}{r^2} + 2\pi r\right) \\[8px] &= -2 \left(-\frac{2V}{r^3} \right) + 2\pi \\[8px] &= \dfrac{4V}{r^3} + 2\pi
\end{align*}

Since $r > 0$, this second derivative $\left( \dfrac{d^2A}{dr^2} = \dfrac{4V}{r^3} + 2\pi\right)$ is always positive $\left(\dfrac{d^2A}{dr^2} > 0 \right)$. Hence this single critical point gives us a minimum:

The minimum area occurs when $r = \sqrt[3]{\dfrac{V}{\pi}}$.

6. Determine the maxima and minima as necessary. Remember to check the endpoints if there are any.

Recall that we found above $h = \dfrac{V}{\pi r^2}$. Hence when $r = \sqrt[3]{\frac{V}{\pi}}$, the cylinder’s height is
\begin{align*}
h &= \dfrac{V}{\pi r^2} \\ \\
&= \dfrac{V}{\pi \left(\sqrt[3]{\frac{V}{\pi}}\right)^2} \\ \\
&= \frac{V}{\pi} \frac{\pi^{2/3}}{V^{2/3}} \\ \\
&= \sqrt[3]{\frac{V}{\pi}}
\end{align*}

Hence we obtain the minimum area for this open-topped can when the cylinder’s height equals its radius, and the dimensions are:
$$ h = r = \sqrt[3]{\dfrac{V}{\pi}} \quad \cmark$$

7. Finally, check to make sure you have answered the question as asked: $x$ or $y$ values, or coordinates, or a maximum area, or a shortest time, or . . . .

The question asked us to specify the cylinder’s dimensions, which we have provided. $\checkmark$

Return to Optimization Problems


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Anonymous

Why choose to write the equation in terms of r and not h?

Matheno

Great question — thanks for asking!

That’s an arbitrary choice: we could have just as easily written the equation for h instead, and then found dA/dh in Step 4. The final answers for the values of r and h that minimize the cost would of course turn out the same as above, as you can verify if you’d like!