Optimization

a) You might be thinking that we are not given enough information. Let’s write down the two equations we know related to cylinders.

$A=2\pi rh+2\pi r^2$

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

We don’t know the area, radius, or height, but we do know the volume, which is 250 ${\text{cm}}^3$ . In all optimization problems, there are always at least two of the following variables:

1. The optimizing variable - This is the variable to be maximized or minimized. It could be profits, volume, area, perimeter, anything.

2. The constraint - This is our limiting variable. This is something that is fixed that will not change in the problem. Usually we are given the value of constraints. What is our constraint in this problem? Volume? Area? Radius? Height?

If you think Area is the constraint... *buzz* so sorry, but no. Area is the optimizing variable, because we want to use as little aluminum as possible. If you picked volume as the constraint, you would be correct, because we are given how many milliliters of soda each can should hold. This is not going to change. Remember, we want to minimize area, and keep the volume constant. What do you think the derivative of the area equation is going to be?

$250=\pi r^{2}h$

We do not want to have to worry about two variables in our area equation, so let’s find h in terms of r by dividing both sides by $\pi r^2$ .

$\frac{250}{\pi r^2}=h$

Now let’s plug this into the Area equation, then take the derivative.

$A=2\cancel{\pi }\cancel{r}\frac{250}{\cancel{\pi }{r}^{\cancel{2}}}+2\pi r^2$

$A=\frac{500}{r}+2\pi r^2$​
Now we can find the derivative of Area and set it equal to zero because this is the quantity we want to minimize. Notice that by setting h in terms of r, we are only differentiating using 1 variable, and do not have to worry about implicit differentiation.

$A^{\prime }=-\frac{500}{r^2}+4\pi r$

$0=-\frac{500}{r^2}+4\pi r$​
$\frac{500}{r^2}=4\pi r$​
$r^3=\frac{500}{4\pi }$​
$r=\sqrt[3]{39.8{\text{ cm}}^3}=3.4\text{ cm}$​

The question asks us to find the dimensions of the can, so let’s knock out height and Area.

$\frac{250}{\pi (3.4{)}^2}=h$

$h=6.9\text{ cm}$​
$A=2\pi (3.4)(6.9)+2\pi (3.4{)}^2$

​
$\mathbf{A}=\mathbf{220}{\text{ cm}}^2$

2. The cost of aluminum is given as a tenth of a penny per ${\text{cm}}^2$ . Our can has an area of 220 ${\text{cm}}^2$ . This is simple math.

$\text{cost per can}=220{\text{ cm}}^2\cdot \frac{\$0.001}{{\text{cm}}^2}=\$0.22$

3. Reality check: See the following image of the proposed can (interactive 6.2.1).