Re: max and min 277
I can't really tell what you are doing. Ordinarily I would not do this, but I feel we are not making progress, so I am going to show you how to work this problem...
Okay, we have the objective function, the surface area of the boiler:
$$S(h,r)=2\pi rh+4\pi r^2$$
and this is subject to the constraint:
$$V=\pi r^2h+\frac{4}{3}\pi r^3$$
Now, if we solve the constraint for $h$, we obtain:
$$h=\frac{3V-4\pi r^3}{3\pi r^2}$$
Now, substituting this into the objective function, we obtain:
$$S(r)=2\pi r\left(\frac{3V-4\pi r^3}{3\pi r^2} \right)+4\pi r^2$$
Simplify:
$$S(r)=2\left(\frac{3V-4\pi r^3}{3r} \right)+4\pi r^2$$
$$S(r)=2Vr^{-1}+\frac{4}{3}\pi r^2$$
Now, differentiate with respect to $r$ and equate the result to zero to obtain the critical value(s):
$$S'(r)=-2Vr^{-2}+\frac{8}{3}\pi r=\frac{2\left(4\pi r^3-3V \right)}{3r^2}=0$$
This implies (for $0<r$):
$$4\pi r^3-3V=0$$
Solving for $r$, we find:
$$r=\sqrt[3]{\frac{3V}{4\pi}}$$
We can see that for any positive $r$, we have $S''(r)>0$, thus we know this critical value is at a minimum.
