Key Concepts and Formulas

• Area of a Circular Sector: $A = \frac{1}{2}r^2\theta$, where $r$ is the radius and $\theta$ is the angle in radians.
• Perimeter of a Circular Sector: $P = 2r + r\theta$, where $r$ is the radius and $\theta$ is the angle in radians.
• Optimization using Derivatives: To find the maximum or minimum of a function, find the critical points by setting the first derivative to zero and use the second derivative test to determine if it's a maximum or minimum.

Step-by-Step Solution

• Step 1: Setting up the equations based on the given information.

We are given that the perimeter of the circular sector is 20 meters. We need to maximize the area of the sector. We can write the equations for the perimeter and area as: $2r + r\theta = 20 \quad \text{(Perimeter Equation)}$ $A = \frac{1}{2}r^2\theta \quad \text{(Area Equation)}$ Explanation: We're translating the problem's conditions into mathematical equations, which is the first step to solving it.

• Step 2: Expressing $\theta$ in terms of $r$ using the perimeter equation.

We want to express the area as a function of a single variable, so we need to eliminate $\theta$. From the perimeter equation, we can solve for $\theta$: $r\theta = 20 - 2r$ $\theta = \frac{20 - 2r}{r} \quad \text{(Equation 3)}$ Explanation: By isolating $\theta$, we can substitute it into the area equation and get a function of $r$ only.

• Step 3: Substituting $\theta$ into the area equation to get $A(r)$.

Substitute the expression for $\theta$ from Equation 3 into the area equation: $A = \frac{1}{2}r^2 \left( \frac{20 - 2r}{r} \right)$ $A(r) = \frac{1}{2}r(20 - 2r)$ $A(r) = 10r - r^2$ Explanation: Now the area is expressed as a function of $r$ only, $A(r)$, which allows us to use calculus to find the maximum area.

• Step 4: Finding the critical points by taking the first derivative of $A(r)$ and setting it to zero.

To find the maximum area, we need to find the critical points of $A(r)$. Take the first derivative of $A(r)$ with respect to $r$: $\frac{dA}{dr} = \frac{d}{dr}(10r - r^2)$ $\frac{dA}{dr} = 10 - 2r$ Set the first derivative to zero: $10 - 2r = 0$ $2r = 10$ $r = 5$ Explanation: Setting the derivative to zero helps us find the points where the function's slope is zero, which are potential maxima or minima.

• Step 5: Verifying that $r=5$ corresponds to a maximum using the second derivative test.

To confirm that $r=5$ corresponds to a maximum, we find the second derivative of $A(r)$: $\frac{d^2A}{dr^2} = \frac{d}{dr}(10 - 2r)$ $\frac{d^2A}{dr^2} = -2$ Since $\frac{d^2A}{dr^2} = -2 < 0$, the critical point $r=5$ corresponds to a local maximum. Explanation: The second derivative test helps us determine if a critical point is a maximum or a minimum. A negative second derivative indicates a maximum.

• Step 6: Calculating the maximum area by substituting $r=5$ into $A(r)$.

Substitute $r=5$ into the area function $A(r) = 10r - r^2$: $A(5) = 10(5) - (5)^2$ $A(5) = 50 - 25$ $A(5) = 25$ Explanation: We substitute the value of $r$ at which we found a maximum into the area equation to find the maximum area.

Common Mistakes & Tips

• Remember to use radians for the angle $\theta$ in the formulas for the area and perimeter of a circular sector.
• Always check the second derivative to confirm whether a critical point is a maximum or a minimum.
• Don't forget to consider the domain of the variables. In this case, both $r$ and $\theta$ must be positive.

Summary

We were given the perimeter of a circular sector and asked to find the maximum area. We set up equations for the perimeter and area, expressed the area as a function of the radius, found the critical point by taking the first derivative and setting it to zero, and then confirmed that the critical point corresponded to a maximum using the second derivative test. Finally, we substituted the value of $r$ that maximizes the area back into the area equation to find the maximum area. The maximum area is 25 sq. m.

Final Answer The final answer is \boxed{25}, which corresponds to option (B).