The problem asks:
A silo (base not included) is to be constructed in the form of a cylinder surmounted by a hemisphere. The cost of construction per square unit of surface area is $7$ times as great for the hemisphere as it is for the cylindrical sidewall. Determine the dimensions to be used if the volume is fixed at $14,000$ cubic units and the cost of construction is to be kept to a minimum. Neglect the thickness of the silo and waste in construction.
Relevant equations include:
Area of the silo = Area of Hemisphere + Area of cylinder
...Which equals:
$A=2\pi r^2+2\pi rh$
I also was able to figure out the cost function - I have to optimize everything in such a way to find out how to spend the least amount of money. The cost function is:
$C=\frac{28000}{r}+\frac{38\pi r^2}{3}$
I know how to find the minimum dimensions via the first and second derivative, but what always trips me up with optimization problems is ascertaining the intervals of interest for the objective function. Obviously we can exclude negative numbers, so it would begin $(0,$ but I don't know what the second part of the interval would be. Any guidance would be much appreciated!
$\endgroup$ 22 Answers
$\begingroup$You are told to minimize $$\tilde C(r,h):=2\pi r(7r+h)$$ under the constraints $$r\geq0,\qquad h\geq0,\qquad\pi r^2\left({2\over3} r+h\right)=V\ ,$$ where $V>0$ is given. The last condition, together with $h\geq0$, enforces $$0<r^3\leq{3V\over2\pi}\ .\tag{1}$$ Using the same condition to eliminate $h$ gives $h={V\over\pi r^2}-{2\over 3}r$, so that we arrive at the the cost function $$C(r)=2\pi r\left({19\over 3}r+{V\over\pi r^2}\right)={38\pi\over3}r^2+{2V\over r}\ .$$ For $r>0$ the right hand side has a unique critical point when $r^3={3V\over38\pi}$. Inspection then tells us that it is decreasing from infinity when $0<r\leq\left({3V\over38\pi}\right)^{1/3}$, and is increasing to infinity thereafter.
Now $(1)$ tells us that the maximal allowed value for $r^3$ is ${3V\over2\pi}$, which is clearly larger than ${3V\over38\pi}$. It follows that the minimum of $C(r)$ in the feasible $r$-interval is $C\bigl({3V\over38\pi}\bigr)$.
Note that in the case of a plastic balloon hemisphere the given factor $7$ might have been $0.1$ instead, and in this case the zero of $C'(r)$ might move outside the interval defined by $(1)$. This means that the optimal choice would then be $r=\left({3V\over2\pi}\right)^{1/3}$, $\>h=0$.
The lesson to learn from this is that you have to obtain a clear overview over the global situation, and should not look at second derivatives at all.
$\endgroup$ $\begingroup$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.
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