What is the area of the largest possible Norman window with a perimeter of 49 feet?

Question

A Norman window has the shape of a semicircle atop a rectangle so that the diameter of the semicircle is equal to the width of the rectangle. What is the area of the largest possible Norman window with a perimeter of 49 feet?

Answer 1

A Norman window is a semicircular window on top of a rectangular window.

let h=height of the window and r=radius of the semicircle

Here is my updated answer and corrected answer.

The diameter of the semicircle is not included in the perimeter. So here are the changes...

P=C+2h+2r (2r=top of the rectangle, h=left and right sides of the rectangle)

seeing that we have a semicircle, C=(1/2)(pi)(d)=pi*r

P=(pi)r+2h+2r

P=2h+(pi)r+2r

49=2h+r(pi+2)

2h=49-r(pi+2)

h=[49-r(pi+2)]/2

A=(1/2)(pi)r²+2rh (area of 1/2 circle + area of rectangle)

substitute the equivalent of h

A=(1/2)(pi)r²+2r[49-(pi)r-2r]/2

A=(1/2)(pi)r²+r[49-r(2+pi)]

A=(1/2)(pi)r²+49r-r²(2+pi)

A=r²[-2-(pi)+(1/2)(pi)]+49r

A=r²[-2-(1/2)(pi)]+49r

use (p,q) where p=r and q=total area

p=-b/2a

p=(-49)/2[-2-(1/2)(pi)]

p=49/(4+pi) which means if r=49/(4+pi), then 2r=98/(4+pi) where 2r=diameter of the circle (semi-circle)

98/(4+3.14159)=98/7.14159=13.722 for the bottom of the rectangle and the diameter of the circle

C=(1/2)(pi)13.722)

C=21.55 (this is 1/2 the circumference)

Now we have to find what is left of the perimeter of 49 feet to determine the sides of the rectangle.

21.55 ft+13.722 ft=35.27 ft, 49-35.27=13.73 ft left for the two sides of the rectangle

13.73/2=6.865 ft for each side

The rectangle is 13.722 ft by 6.865 ft

Now find the area.

A=(1/2)(pi)(6.865)²+(13.722)(6.865)

A=(1/2)(3.14159)(47.128)+(13.722)(6.865)

A=(1.5708)(47.128)+(13.722)(6.865)

A=74.02+94.20

A=168.22 square feet

Note: Some pictures of Norman windows had a semicircle on top of a rectangle so I included the diameter of the circle (the top of the rectangle) in the perimeter and in the calculations. The window in the problem apparently did not include the diameter as part of the perimeter.

 

Answer 2

Alison, the goal to solving this problem is to come up with the area of the window as a function of the diameter of the semicircle, take the derivative of the function, set it to equal to zero, solve it for the diameter, and substitute into the original equation to determine the maximum area. Start with writing an equation for the perimeter.

 

P=49=1/2*pi*D+D+2L 

 

D=the diameter of the semicircle and let it become the independent variable

L=the length of a side of the rectangular part of the window.

 

A(D)=L*D+1/2*pi*(D/2)^2

 

Solve for L in the perimeter equation.

 

2L=49-D(1/2*pi+1) Note: I just subtracted the first two terms from both sides of Perimeter equation.

 

This implies L=(49-D(1/2*pi+1)/2. Hence the area function becomes:

 

A(D)=(49D-D^2*(1/2*pi+1)/2+1/8*pi*D^2 With some algebraic manipulation the function becomes.

 

A(D)=49/2*D-1/4*pi*D^2-1/2*D^2+1/8*pi*D^2 = 49/2*D-1/8*pi*D^2-D^2/2

A(D)=49/2*D-1/2*D^2*(1/4*pi+1)

 

Take the derivative and set it to equal to zero.

 

A'(D)=49/2-2D/2(1/4*pi+1)=49/2-D(1/4*pi+1)=0 This implies:

 

D=24.5/(1/4*pi+1) Note: 24.5=49/2. Just plug this value for D into the equation for area to get the maximum value.

 

Hope this helps.

 

 

Answer 3

So this is an optimization problem.  First lets set up what we know

 

We know total perimeter is 16 and the diameter of the semicircle is the width of the rectangle.  π=Pi

 

So Perimeter=49=2L+W+1/2(πW)

 

The first two components (2L+W) come from the sides of the rectangle and the 1/2(πW) is the circumference(or perimeter) of the semicircle which in total gives you the perimeter of total window or 49.

 

Next we need an equation for the area of the window.

 

A = LW+1/2(π)*(1/2(W))²

 

The first component is the area of the rectangle (LW) and the second is the area of the semicircle 1/2(π)*(1/2(w))²

 

So now you have to Maximize the area with the constraint of the perimeter.

 

In order to do that you must solve the perimeter equation for L. 

Then plug that L into the L in the area equation.

Then take derivative of area with respects to W.  Set the result to zero and solve for W.  You then plug that W into your Area equation to get the max area.

 

I felt the set up was likely the harder part of this problem but would be happy to explain the last part more in depth if you need