Answer: (a) \(P=2r+\frac{20}{r}\); (b) \(r=\sqrt{10}\); (c) minimum; (d) \(\theta=2\).

Answer: (a) \(P=2r+\frac{20}{r}\); (b) \(r=\sqrt{10}\); (c) minimum; (d) \(\theta=2\).

The sector area is

\(\frac12r^2\theta=10.\)

Thus

\(\theta=\frac{20}{r^2}.\)

The perimeter is two radii plus the arc length:

\(P=2r+r\theta=2r+r\left(\frac{20}{r^2}\right)=2r+\frac{20}{r}.\)

Differentiate:

\(\frac{dP}{dr}=2-\frac{20}{r^2}.\)

For a stationary value,

\(2-\frac{20}{r^2}=0,\)

so \(r^2=10\). Since \(r\gt 0\),

\(r=\sqrt{10}.\)

Now

\(\frac{d^2P}{dr^2}=\frac{40}{r^3}.\)

Since \(r\gt 0\), this is positive, so the stationary value is a minimum. Finally,

\(\theta=\frac{20}{r^2}=\frac{20}{10}=2.\)