# A sector of a circle whose radius is r and whose angle is theta has a fixed perimeter P. How do you find the values of r and theta so that the area of the sector is a maximum?

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To find the values of ( r ) and ( \theta ) so that the area of the sector is a maximum, we first need to express the area of the sector in terms of ( r ) and ( \theta ). Then, we can use calculus to maximize the area.

The perimeter of the sector, ( P ), is given by ( P = 2r + r\theta ).

The area of the sector, ( A ), is given by ( A = \frac{1}{2} r^2 \theta ).

To maximize the area, we can differentiate the area function ( A ) with respect to ( \theta ), set the derivative equal to zero, and solve for ( \theta ). Then, we can substitute the value of ( \theta ) back into the expression for the perimeter to find the corresponding value of ( r ).

Differentiating ( A = \frac{1}{2} r^2 \theta ) with respect to ( \theta ), we get ( \frac{dA}{d\theta} = \frac{1}{2} r^2 ).

Setting ( \frac{dA}{d\theta} ) equal to zero, we have ( \frac{1}{2} r^2 = 0 ).

Solving for ( \theta ), we find that ( \theta = 0 ).

Substituting ( \theta = 0 ) into the expression for the perimeter, ( P = 2r + r\theta ), we get ( P = 2r ).

So, to maximize the area of the sector, ( \theta ) should be 0, and ( r ) should be half of the fixed perimeter ( P ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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