Question

A church window consisting of a rectangle topped by a semicircle is to have perimeter p . find the radius of the semicircle if the area of the window is to be maximum.

Answer 1

Answer:

$r=\dfrac{p}{4+\pi}$

Step-by-step explanation:

Let r be the radius of the semicircle, then 2r is the width of the rectangle. Let y be the length of the rectangle. The perimeter of the window consists of two lengths, one width and length of semicircle, then

$p=2y+2r+\pi r.$

Express y:

$y=\dfrac{p-2r-\pi r}{2}.$

The area of the window is

$A=y\cdot 2r+\dfrac{1}{2}\pi r^2.$

Substitute y into the area expression:

$A=\dfrac{p-2r-\pi r}{2}\cdot 2r+\dfrac{1}{2}\pi r^2=pr-2r^2-\pi r^2+\dfrac{1}{2}\pi r^2=pr-2r^2-\dfrac{1}{2}\pi r^2.$

Find the derivative A':

$A'=p-4r-\pi r.$

Equate A' to 0:

$p-4r-\pi r=0\Rightarrow r=\dfrac{p}{4+\pi}.$

When $r<\dfrac{p}{4+\pi},$ then $A'>0$ and the function A is increasing, when $r>\dfrac{p}{4+\pi},$ then $A'<0$ and the function A is decreasing. This means that at point $r=\dfrac{p}{4+\pi}$ the function A takes it maximal value (the area is maximal).

Answer 2

Function assigns the value of each element of one set to another element of another set. At point, $r = \dfrac{P}{4+\pi}$ , function A will be maximum.

What is a Function?

A function assigns the value of each element of one set to the other specific element of another set.

Assume that the radius of the semicircle is r, while the length of the rectangle is L. therefore, 2r or the diameter of the width of the rectangle.

Now, the perimeter of the window consists of two lengths, one width, and length of the semicircle, therefore,

$\rm Perimeter, P = 2L + 2r + \pi r$

The length of the window can be written as,

$L=\dfrac{P-2r-\pi r}{2}$

The area of the window can be written as,

$\text{Area of the window}=(L \times 2r)+ \dfrac{1}{2}\pi r^2$

Substituting the value of the length of the window,

$\text{Area of the window}=(\dfrac{P-2r-\pi r}{2}\times 2r)+ \dfrac{1}{2}\pi r^2$

                               $=Pr-2r^2-\pi r^2+ \dfrac{1}{2}\pi r^2\\\\= Pr-2r^2- \dfrac{1}{2}\pi r^2$          

     

Finding the derivative of the Area, A with respect to the radius, r.

 $A' = P-4r-\pi r$

Equating the derivative with 0,

$P-4r-\pi r = 0\\\\r = \dfrac{P}{4+\pi}$

Thus, if $r < \dfrac{P}{4+\pi}$  then A'>0 therefore, the function A is increasing, and when $r > \dfrac{P}{4+\pi}$  then A'<0 therefore, the function A is decreasing.

Hence, this means that at the point $r = \dfrac{P}{4+\pi}$ the function A will be maximum.

Learn more about Function:

https://brainly.com/question/5245372