Answer: \(h^2=\frac{16}{r^2}-r^2\); \(V=\frac{\pi}{3}\sqrt{16r^2-r^6}\); \(r=\frac{2}{\sqrt[4]{3}}\).

(a) The curved surface area is \(4\pi\), so

\(\pi rl=4\pi.\)

Cancel \(\pi\):

\(rl=4.\)

Thus

\(l=\frac4r.\)

For a right circular cone, the radius, height and slant height form a right-angled triangle, so

\(l^2=r^2+h^2.\)

Hence

\(h^2=l^2-r^2.\)

Substitute \(l=\frac4r\):

\(h^2=\left(\frac4r\right)^2-r^2.\)

Therefore

\(h^2=\frac{16}{r^2}-r^2.\)

(b) Since \(V=\frac13\pi r^2h\),

\(V=\frac13\pi r^2\sqrt{\frac{16}{r^2}-r^2}.\)

Rewrite the expression inside the square root:

\(\frac{16}{r^2}-r^2=\frac{16-r^4}{r^2}.\)

Since \(r\gt 0\),

So

\(V=\frac{\pi}{3}r\sqrt{16-r^4}.\)

This can be written as

(c) To maximise \(V\), it is enough to maximise the expression inside the square root:

\(F(r)=16r^2-r^6.\)

Differentiate:

\(F'(r)=32r-6r^5.\)

Set this equal to zero:

\(32r-6r^5=0.\)

Factorise:

\(2r(16-3r^4)=0.\)

Since \(r\gt 0\),

\(16-3r^4=0.\)

Thus

\(r^4=\frac{16}{3}.\)

Taking the positive fourth root,

This gives the maximum volume, since \(V\) is zero at the endpoints of the possible interval and positive inside it.