Answer: \(A=\frac{91}{8}x^2-68x+132\); minimum area \(=\frac{2764}{91}\approx30.4\text{ cm}^2\).

We start with the main method. Use the measurements and relationships shown in the diagram, then translate them into algebraic or trigonometric equations. Use the right-triangle information to find the angle in radians, then apply arc length and sector area formulae.

(a) The perimeter is made from the three arcs and the three exposed straight radial parts.

The arc lengths are

\(\frac12x,\quad 2(x+2),\quad y.\)

The exposed straight lengths are

\(x,\quad 2,\quad x+2-y,\quad y.\)

So the perimeter is

\(\frac12x+2(x+2)+y+x+2+(x+2-y)+y.\)

Since the perimeter is \(24\),

\(\frac92x+y+8=24.\)

Thus

\(y=16-\frac92x.\)

The total area is the sum of the three sector areas:

\(A=\frac12x^2\left(\frac12\right)+\frac12(x+2)^2(2)+\frac12y^2(1).\)

So

\(A=\frac14x^2+(x+2)^2+\frac12\left(16-\frac92x\right)^2.\)

Expanding gives

\(A=\frac14x^2+x^2+4x+4+\frac12\left(256-144x+\frac{81}{4}x^2\right).\)

Therefore

\(A=\frac{91}{8}x^2-68x+132.\)

(b) Differentiate:

\(\frac{\mathrm dA}{\mathrm dx}=\frac{91}{4}x-68.\)

For a minimum, set this equal to zero:

\(\frac{91}{4}x-68=0.\)

Hence

\(x=\frac{272}{91}.\)

Since

\(\frac{\mathrm d^2A}{\mathrm dx^2}=\frac{91}{4}\gt0,\)

this gives a minimum.

The minimum area is

\(A=\frac{91}{8}\left(\frac{272}{91}\right)^2-68\left(\frac{272}{91}\right)+132 =\frac{2764}{91}.\)

So the minimum possible area is approximately

\(30.4\text{ cm}^2.\)

This completes the solution and gives the required result.