Question:

Step 1: Set up the perimeter constraint for the sector. The perimeter of a sector is given by the sum of two radii and the arc length. Let $r$ be the radius and $\theta$ be the angle in radians. The perimeter $P$ is given as 20 metres. $$ P = r + r + r\theta = 2r + r\theta $$ Given $P = 20$ m: $$ 2r + r\theta = 20 $$ Factor out $r$: $$ r(2 + \theta) = 20 $$ Express $r$ in terms of $\theta$: $$ r = \frac{20}{2 + \theta} $$ Step 2: Express the area of the sector as a function of $\theta$. The area $A$ of a sector with radius $r$ and angle $\theta$ (in radians) is given by: $$ A = \frac{1}{2}r^2\theta $$ Substitute the expression for $r$ from Step 1 into the area formula: $$ A = \frac{1}{2}\left(\frac{20}{2 + \theta}\right)^2\theta $$ $$ A = \frac{1}{2} \cdot \frac{400}{(2 + \theta)^2}\theta $$ $$ A = \frac{200\theta}{(2 + \theta)^2} $$ Step 3: Differentiate the area function with respect to $\theta$. To find the maximum area, we need to find the critical points by differentiating $A(\theta)$ with respect to $\theta$ and setting the derivative to zero. We use the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$. Let $u = 200\theta$ and $v = (2 + \theta)^2$. Then $u' = 200$ and $v' = 2(2 + \theta) \cdot 1 = 2(2 + \theta)$. $$ \frac{dA}{d\theta} = \frac{200(2 + \theta)^2 - 200\theta \cdot 2(2 + \theta)}{((2 + \theta)^2)^2} $$ $$ \frac{dA}{d\theta} = \frac{200(2 + \theta)^2 - 400\theta(2 + \theta)}{(2 + \theta)^4} $$ Factor out $200(2 + \theta)$ from the numerator: $$ \frac{dA}{d\theta} = \frac{200(2 + \theta)[(2 + \theta) - 2\theta]}{(2 + \theta)^4} $$ Simplify the expression: $$ \frac{dA}{d\theta} = \frac{200(2 - \theta)}{(2 + \theta)^3} $$ Alternatively, as in the original solution's simplification: $$ \frac{dA}{d\theta} = \frac{200(2+\theta)^2 - 400\theta(2+\theta)}{(2+\theta)^4} $$ Divide numerator and denominator by $(2+\theta)$: $$ \frac{dA}{d\theta} = \frac{200(2+\theta) - 400\theta}{(2+\theta)^3} $$ Expand and simplify the numerator: $$ \frac{dA}{d\theta} = \frac{400 + 200\theta - 400\theta}{(2+\theta)^3} $$ $$ \frac{dA}{d\theta} = \frac{400 - 200\theta}{(2+\theta)^3} $$ Step 4: Find the critical point by setting the derivative to zero. Set $\frac{dA}{d\theta} = 0$ to find the value of $\theta$ that maximizes the area: $$ \frac{400 - 200\theta}{(2 + \theta)^3} = 0 $$ For the fraction to be zero, the numerator must be zero: $$ 400 - 200\theta = 0 $$ $$ 200\theta = 400 $$ $$ \theta = 2 \text{ radians} $$ Step 5: Calculate the corresponding radius $r$. Substitute the value $\theta = 2$ radians back into the expression for $r$ from Step 1: $$ r = \frac{20}{2 + \theta} $$ $$ r = \frac{20}{2 + 2} $$ $$ r = \frac{20}{4} $$ $$ r = 5 \text{ metres} $$ Step 6: Calculate the maximum area. Substitute the values of $r = 5$ m and $\theta = 2$ radians into the area formula: $$ A = \frac{1}{2}r^2\theta $$ $$ A = \frac{1}{2}(5)^2(2) $$ $$ A = \frac{1}{2}(25)(2) $$ $$ A = 25 \text{ m}^2 $$ Step 7: Verify that this value corresponds to a maximum. To confirm that this is a maximum, one can perform the second derivative test. The second derivative $\frac{d^2A}{d\theta^2}$ would be evaluated at $\theta = 2$. If $\frac{d^2A}{d\theta^2} < 0$, it confirms a local maximum. For this problem, it is a standard result that for a fixed perimeter, the maximum area of a sector occurs when $\theta = 2$ radians. The maximum area of the sector is $25 \text{ m}^2$. The final answer is $\boxed{\text{25}}$.