MHB What is the maximum area of the circular sector with a perimeter of 12 cm?

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The discussion focuses on finding the maximum area of a circular sector with a perimeter of 12 cm, expressed as functions of both the central angle θ and the radius r. The area of the sector is initially expressed incorrectly, but the correct function for A(θ) is given as A(θ) = (72θ)/(2 + θ)², which is not quadratic. For A(r), the correct expression is A(r) = 6r - r², which is identified as a quadratic function. Participants seek clarification on the steps to maximize the area and confirm the relationships between perimeter and area formulas. The conversation emphasizes understanding the correct expressions and their implications for maximizing the sector's area.
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The given picture shows a circular sector with radius r cm and central angle θ (radian measure). The perimeter of the sector is 12 cm.

1. Express the area A of the sector as a function of θ. Is this a quadratic function?

2. Express the area A of the sector as a function of r. Is this a quadratic function?

3. For which value of r is the area A a maximum? What is the corressponding value of θ in this case?

Part 1

I know the area of the sector formula is A = (1/2) r^2 θ.

I simply thought the answer must be A(θ) = (1/2) r^2 θ but then I said no because the radius square should not be as part of the answer. The question is asking to express the area A of the sector as a function of theta not a function of r and θ.

The book's answer is A(θ) = (72 θ)/(2 + θ)^2. Of course, this is not a quadratic function. Where did this fraction come from? Where did 72 and 6 come from?

Part 2

Again, I quickly jumped to conclusion since I know the area of a sector formula to be A = (1/2) r^2 θ.

I said, ok, the answer must be A(r) = (1/2) r^2 θ. There it is. It is written in terms of r. I am wrong, of course. You see, θ should not be part of the answer if we must express the area A as a function of r not a function of r and θ.

The book's answer is A(r) = 6r - r^2. I recognize this to be a quadratic function. Where did 6 come from?

Part 3

I need the steps for Part 3 that will guide me to the answer. Can someone provide me with the steps?

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We know the perimeter $P$ is:

$$P=2r+r\theta=(2+\theta)r\tag{1}$$

The area is:

$$A=\frac{1}{2}r^2\theta\tag{2}$$

(1) gives is a way to relate $r$ and $\theta$. So:

1.) solve (1) for $r$, and substitute for $r$ into (2) to get $A(\theta)$. Now, you will also have the parameter $P$ present, but it has been given, so you can just plug that in once you get the formula. What do you get?

2.) solve (1) for $\theta$, and substitute for $\theta$ into (2) to get $A(r)$. Now, as before, you will also have the parameter $P$ present, but it has been given, so you can just plug that in once you get the formula. What do you get?
 
You say solve 1 for r and plug for r into 2.
Then you said to solve 1 for theta and plug for theta into 2.

Do you mean that 1 here represents the perimeter of the sector formula and 2 the area of the sector formula?

If so, I will try later tonight after work.
 
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