Probem30+Anders,+Mike+J.

optimizing eq A= lw+[(d/2)pi^2]/2 d=w Domain 0I could be wrong here, but following your work I get A' = (30-2d - 2π)/2 + (πd)/ 2 , which gives a different value for d. 0=[30-2d-pid/2]+pi^2/4 d=[30+pi^2/2]/[2+pi] d=6.79 this is one contender for max area

lim of A as d->0= 0 no good lim of A as d-> 30= -632.84 no good

A(6.79)= 59.34 when you plug 6.79 back into the P equation you get l to equal 6.27 so they are the dimensions to maximize area.

Mike, I agree with your work below except that the perimeter of the whole window is 30 -- so according to your work, l + 2w + (1/2) (2 ∏l)=30

The window would look like a rectangle with a semicircle on top of it, sharing the length of the top side of the rectangle. This means that the diameter of this semicircle is equal to the length of the horizontal sides of the rectangle.  To find the maximum area of this window, we set up the equation combining the area of the rectangle with that of the semicircle: A= Lw + 1/2(π)(L/2)^2 This simplifies to A= Lw + (π/8)L^2

We know that the perimeter of the rectangle must equal 30, so we can write 2L +2w = 30. We can then solve for w and get w= 15-L. If we substitute this value into our area equation, we can eliminate the L variable: A= L(15-L) + (π/8)L^2. This simplifies to A= 15L - L^2 + (π/8)L^2. We can then find the derivative of A and set that equal to zero.

A' = 15 - 2L + (π/4)L = 0 15 = 2L - (π/4)L 15 = 1.215L w = 15 - L
 * L = 12.35**
 * w = 2.65**

When we plug in these values into our first Area equation we get A = 92.62. We should check to see that the endpoints of our domain do not give a greater value for the area of the window. Since the domain is (0,30), we plug in 0 for w and 15 for L (because there are 2 lengths) and get A = 88.36.

Therefore the dimensions of the window that would admit the most light would be L = 12.35 and w = 2.65.