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!
Best Answer
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.