# Feasible Domain Optimization Problems Homework

There are a couple of ways to approach a problem like this. In addition to the formula $A=2\pi r^2+2\pi rh$, you have two constraints: $r > 0$ and $h \geq 0$.

Now you want to minimize $$C(r)=\frac{28000}{r}+\frac{38\pi r^2}{3}.$$

It happens that you really need to apply the first constraint to solve the problem, because $C(r)$ has no minimum on $(-\infty,0) \cup (0,\infty)$, which is its domain. But $C(r)$ does have a minimum at a unique point $r_1$ in $(0,\infty)$, and if we compute $h$ at $r_1$ we will find that $h \geq 0$, so we accept $r_1$ as the value of $r$ that minimizes $C(r)$.

The shady part of this method is that we know that not all values of $r$ in $(0,\infty)$ could even be considered as possible solutions, because for very large values of $r$, when you solve for $h$ in $A=2\pi r^2+2\pi rh$ you get $h < 0$, contradicting one of your constraints. There is, in fact, a maximum value $r_\max$ of $r$ that can be achieved within the given constraints. The cost function is not defined outside the interval $(0,r_\max]$ since you can't build a silo for any other $r$. What I did above was to guess that $r_\max$ is greater than the $r$ that minimizes $C(r)$ over all of $(0,\infty)$, and after finding that value of $r$, I confirmed that my guess was correct.

If I omitted that last step from my solution, I would still present the correct value of $r$ to minimize cost, but I would not really have *solved* the problem because I would not have *proved* that this was the correct value of $r$.

What your homework site seems to be demanding is that you don't engage in such guesswork, even if you can justify it later. I suppose the site wants you to use the constraint $h\geq 0$ to find the value of $r_\max$ so that you can give the interval $(0,r_\max]$ explicitly.

So the question is, how large can you make $r$ while still having $A=2\pi r^2+2\pi rh$ and $h \geq 0$? I think it's "obvious" that the largest possible value of $r$ occurs when $h=0$, but you might want to think about it a bit.

answered Oct 29 '16 at 18:13

This example is from Paul's Online Notes for Calc I.

You have $500$ feet of fencing material and you want to enclose a field with a fence. A building is on one side of the field (and so won't need any fencing). Determine the dimensions of the field that will enclose the largest area.

Maximize: $\;\;\;A=xy$

Constraint: $\;\;\;500=x+2y$

$x=500-2y \implies A(y)=(500-2y)(y) \implies A(y)=500y-2y^2$

Then, for determining the interval, he states:

Now we want to find the largest value this will have on the interval [$0,250]$. Note that the interval corresponds to taking $y=0$ (i.e. no sides to the fence) and $y=250$ (i.e. only two sides and no width, also if there are two sides each must be $250$ ft to use the whole $500$ft).

**My Questions:**

$\textbf{1.}\;\;$ I don't understand his explanation for determining the interval...why is he considering no sides to the fence, which he labels as $y=0$, and only two sides and no width, which he labels as $y=250$? It seems like these numbers come from $2y(250-y) \implies y=0, y=250$, but I still don't understand why this is the case. How can you ignore dimensions when determining the domain?

$\textbf{2.}\;\;$ How do you know that the largest area will be rectangular? Why doesn't he choose to maximize a circle or square, for example?

Thanks.

calculusfunctionsoptimization

asked Jan 14 '14 at 23:21

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