Optimization

Optimization is at the heart of calculus’ application to the world around us. We have learned about extremal points and now we get to put them into practice by considering real world situations. Additionally, we will focus on reducing the dimensionality of problems with constraint equations.

1. What is optimization?

Often, we can express a situation or problem we wish to solve (perhaps somewhat simplified) in terms of a differentiable function, $f(x)$, or, perhaps, $f(x,y,z)$. We may become interested in the maxima and minima of this model or to related constraint, $c(x)$. These extremal solutions, of course, correspond to either $f'(x) = 0$ or $c'(x) = 0$. From here, they trickle into all sorts of other areas of modern mathematics including game theory and decision theory.

There are two sneaky ideas here:

The full power of calculus is only available in calculus of one variable. Often, we look for symmetry, dependencies, or other ideas to reduce the number of apparent variables to just one.
We may be able to treat a variable as parameter or a constant to optimize along that parameter if we are certain of our optimization. For instance, if a cost minimization is all important than time may not be a factor. Similarly, if time is the most most important constraint then we would optimize along that parameter.

Now, it turns out the reality of the situation is often more complex and then we might resort to something along the lines of LanGrange multipliers in a multivariable setting.

I’d like to emphasize here that in calculus of one variable, we can leverage the full power of the limit precisely because there are only two directions from which we can approach a particular limit. In every dimension higher than 1, there are infinitely many directions and paths through which I approach to get to the point I am interested in. This is incorporated into the limit in higher dimensions and as a result differentiation and integration work out much the same way, however, some of our most beloved theorems either do not work or become much more difficult to apply.

Recall the extreme value theorem. This is among our most powerful results in calculus of one variable. It is true in every finite dimension, but becomes difficult to apply because there are necessarily infinite many boundary points. Reflect on this when you see LaGrange multipliers!

Worked Problems

1. Suppose I wish to construct a cylindrical tank with a volume 36 liters. I want to do this in such a way as to minimize the material used to construct the tank. What dimensions should the tank be?

We are given the volume of the tank and, moreover, we know the formula for the volume of a cylinder. We will use both of these facts in tandem. Additionally, we wish to convert volume given in liters into a measurement that is more conducive to a dimension we can fabricate with: cubic centimeters.

$V = 36 L = 36000 mL = 36000 cm^3 = \pi r^2h$

I realize that I will eventually minimize the surface area as that directly corresponds to the amount of material needed to produce this tank. I rearrange the above equation to solve for height because I realize my surface area equation will have S = surface area, h = height, and r = radius and I wish to reduce this to two variables.

$\frac{36000}{\pi r^2} = h$

$S = 2\pi r^2 + 2\pi r h$

$S = 2\pi r^2 + \frac{72000}{r}$

$S’ = 4\pi r – \frac{72000}{r^2}$

$0 = 4\pi r – \frac{72000}{r^2}$

$0 = 4\pi r^3 – 72000$

$\sqrt[3]{\frac{18000}{\pi}} = r \approx 17.894 \text{ cm}$

$h = \frac{36000}{\pi (\sqrt[3]{\frac{18000}{\pi}})^2} \approx 37.788 \text{ cm}$

Technically, we should also justify that the point we found was a minimum. Let us consider

$S^{”} = 4\pi + \frac{144000}{r^3}$ We notice that S” will always be positive for $r > 0$ thus the $r$ we produced must be a minimum.

2. Suppose I construct a rectangular box out of varying materials. The top of the container costs $\text{\$ 10}$ per square foot to fabricate and the sides and bottom of the box costs $\text{\$ 4}$ per square foot. The volume of the box will be 32 $ft^3$. The length needs to be 4 times the width. What is the minimum cost of the container.

$v = 32 = hwl$

$l = 4w$

$32 = 4hw^2$

$\frac{8}{w^2} = h$

$C(h, w, l) = 10wl + 2 \cdot 4 wh + 2 \cdot 4 lh + 3wl$

$C(h, w, l) = 13wl + 8wh + 8lh$

$C(h, w) = 13w(4w) + 8wh + 8(4w)h$

$C(h, w) = 52w^2 + 8wh + 32wh$

$C(w) = 52w^2 + 8w(\frac{8}{w^2}) + 32w(\frac{8}{w^2})$

$C(w) = 52w^2 + \frac{320}{w}$

$C'(w) = 104w – \frac{320}{w^2} = 0$

$0 = 104w^3 – 320$ Note that we multiplied through by $w^2$ here, but we are only able to do this because we know with certainty that w is not zero. If it were, then I would not have a box!

$\sqrt[3]{\frac{320}{104}} = w \approx 1.454 \text{ ft} \rightarrow C = \text{\$} 330.02$

3. You want to construct a box with no top with the maximum volume by cutting squares pieces from the corners of a rectangular sheet of carboard, folding the resulting sides up and taping them together as given below. The sheet is L long and W wide. Determine the height of the box.

$V(h,w,l) = hwl$

Let x be the side of one of the squares. This will also be the height for which we are solving.

$h = x$

$l = L – 2x$

$w = W – 2x$

$V(x) = x(W-2x)(L-2x)$

$V(x) = xWL -2x^2W – 2x^2L+ 4x^3$

$V'(x) = WL – 4x(W+L)+ 12x^2$

$0 = WL – 4x(W+L)+ 12x^2 \rightarrow$

$x = \frac{4(W+L) \pm \sqrt{16(W+L)^2 – 4(12)(WL)}}{24} =\frac{(W+L) \pm \sqrt{W^2 + L^2 – WL}}{6}$

Now, it is not clear which, if either solution is a maximum, so we may consider the second derivative test here:

$V”(x) = -4(W+L) + 24x < 0$ if x is a maximum.

$x < \frac{W+L}{6}$ which tells us that only

$x =\frac{1}{6}(W+L – \sqrt{W^2 + L^2 – WL})$ is a solution.