Answer: \(P=2r+\frac{50}{r}\), and the minimum value of \(P\) is \(20\).

For a sector, use radians throughout with arc length \(s=r\theta\) and area \(A=\frac12r^2\theta\).

(a) The area of the sector is

\(\frac12r^2\theta=25.\)

So

\(\theta=\frac{50}{r^2}.\)

The perimeter of the minor sector is two radii plus the arc length:

\(P=2r+r\theta.\)

Substitute \(\theta=\frac{50}{r^2}\):

\(P=2r+r\left(\frac{50}{r^2}\right) =2r+\frac{50}{r}.\)

(b) Differentiate \(P\) with respect to \(r\):

\(\frac{\mathrm dP}{\mathrm dr}=2-\frac{50}{r^2}.\)

At a stationary point,

\(2-\frac{50}{r^2}=0.\)

So

\(r^2=25,\)

and since \(r\gt 0\),

\(r=5.\)

To check that this is a minimum,

\(\frac{\mathrm d^2P}{\mathrm dr^2}=\frac{100}{r^3}.\)

At \(r=5\), this is positive, so \(P\) has a minimum.

The minimum value is

\(P=2(5)+\frac{50}{5}=10+10=20.\)

Therefore the result matches the required answer.