A Norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular window (see figure).

Question

A Norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular window (see figure). Find the dimensions of a Norman window of maximum area if the total perimeter is 18 feet.

Answer 1

Let L be the height of the rectangular part of the window and W be the width, which is also the diameter of the semicircular part of the window. The perimeter P of the window is therefore P = L+W+L+(pi/2)W. Setting this equal to 18 and solving for L is terms of W gives L = 9-(W/4)*(2+pi).

The area A of the window is the rectangular part plus the semicircular part, or A = L*W+(pi/2)*(W/2)*(W/2). Substituting in the expression for L gives us an area A solely in terms of the width W: A = 9W - (W*W)[(1/2)+(pi/8)]. The maximum area occurs when the derivative is 0, or when dA/dW = 9 - (2W)[(1/2)+(pi/8)] = 0. Solving for W gives W = 36/(4+pi), and plugging this into the expression for L gives L = 18/(4+pi). As a check, you can confirm that the perimeter P = L+W+L+(pi/2)W does equal 18 when you plug in W and L. The area of the window A is A = L*W+(pi/2)*(W/2)*(W/2) which, after some simplification, is 162/(4+pi)

Answer 2

Though there is not a figure shown, I will make assumptions. I will assume that the radius of the upper semicircle is r, and the height of the lower rectangular window is h. In this way, the width of that rectangular window is 2r to match the top that is the diameter of the semicircle. The area of this window is the area of a semicircle plus the area of the rectangle.

A=1/2πr²+2rh

Then, we know that the perimeter of this window is 18 feet. For this, we will use 3 of the 4 sides of the rectangle, 2h+2r, and for the curve, that is half of the circumference of the circle: 1/2(2πr), or simply πr

Thus, 2h+2r+πr=18

We have two equations, and we need to maximize the area one, so for that, we will solve the perimeter equation for h, and in that way, be able to make the area equation in terms of only r

So, we subtract all but 2h from both sides of the perimeter equation, which gives us 2h=18-2r-πr.

Then, divide all terms by 2, so h=9-r-πr/2

Now, we can replace h in the area equation with this, and we have A(r)=1/2πr²+2πr(9-r-πr/2)

Distribute, and this is A(r)=1/2πr²+18πr-2πr²-π²r²

We could combine like terms, but it wouldn't give us a whole lot, and it is a little strange. So, to maximize the area, we need to find where A'(r)=0, so we do the derivative of this. Remember, π is not a variable. It is a constant, so we are finding the derivative with respect to only r.

A'(r)=πr+18π-4πr-2π²r=0

Now, it makes sense to combine like terms, and we have 18π-3πr-2π²r=0

Now, we want to get all terms with r on one side of the equation, so I will add them to get 18π=3πr+2π²r

If we factor out an r, we get 18π=(3π+2π²)r

Then, divide, and r=18π/(3π+2π²)

We can use π=3.14 to approximate this, and that gives us that r=1.94 feet.

Then, since h=9-r-πr/2, we can plug in r=1.94, and h=9-1.94-(3.14*1.94/2)=5.95 feet.

Thus, the dimensions of the window will be r=1.94 ft, h=5.95 ft, and the width, using that radius, is 3.88 feet