Answer: (a) \(h=\frac{300-3\pi r^2}{2\pi r}\); (b) shown; (c) maximum volume \(437\text{ cm}^3\) approximately.

Answer: (a) \(h=\frac{300-3\pi r^2}{2\pi r}\); (b) shown; (c) maximum volume \(437\text{ cm}^3\) approximately.

(a) The outside surface area consists of the curved surface of the cylinder, the curved surface of the hemisphere, and the circular base.

The curved surface area of the cylinder is

\(2\pi rh.\)

The curved surface area of the hemisphere is half the surface area of a sphere:

\(2\pi r^2.\)

The base contributes

\(\pi r^2.\)

So the total surface area is

\(2\pi rh+3\pi r^2=300.\)

Rearranging,

\(2\pi rh=300-3\pi r^2.\)

Hence

\(h=\frac{300-3\pi r^2}{2\pi r}.\)

(b) The volume of the object is the volume of the cylinder plus the volume of the hemisphere:

\(V=\pi r^2h+\frac23\pi r^3.\)

Substitute the expression for \(h\):

\(V=\pi r^2\left(\frac{300-3\pi r^2}{2\pi r}\right)+\frac23\pi r^3.\)

This simplifies to

\(V=\frac r2(300-3\pi r^2)+\frac23\pi r^3.\)

Therefore

\(V=150r-\frac32\pi r^3+\frac23\pi r^3.\)

So

\(V=150r-\frac56\pi r^3.\)

(c) Differentiate:

\(\frac{dV}{dr}=150-\frac52\pi r^2.\)

At a stationary point,

\(150-\frac52\pi r^2=0.\)

Thus

\(r^2=\frac{60}{\pi}, \qquad r=\sqrt{\frac{60}{\pi}}.\)

Substituting this into \(V=150r-\frac56\pi r^3\) gives

\(V=437\ldots\)

Therefore the maximum volume is

\(437\text{ cm}^3\)

to 3 significant figures.