calculus optimization question?

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 41 feet? does neone know how to do this question?

Answer 1

Let x = width, y = length of rectangle. Then because the perimeter is 41, we know

41 = pi x + x + 2y = x(pi + 1) + 2y
y = 1/2(41 - x(pi + 1))

Area of window

= 1/8 pi x^2 + xy
= 1/8 pi x^2 + x(1/2(41 - x(pi+1)))
= 1/2(41) x - (3/8 pi +1/2) x^2

Differentiate with respect to x, and we get:

1/2(41) - (3/4 pi + 1) x

Set this to 0 for maximum/minimum:

0 = 1/2(41) - (3/4 pi + 1) x
x = 1/2(41) / (3/4 pi + 1)
x = 6.108 approximately
y = 7.852 approximately

Area = 1/8 pi x^2 + xy = 62.61 approximately

Addendum: Grl is wrong
Addendum 2: Wal C has made good effort but he needs to check his figures

Answer 2

Draw a picture.

label the length of the rectangle x and the diameter(width) d

set up an equation for the area of the rectangle (A=xd)

set up an equation for the perimeter meter of the whole window (pi*r+2x+d)=41

substitute .5d for the r and solve for either variable. I recommend d which gives you d=(41-2x)/(1+.5pi)

Use that to substitute for d in the area formula. so, A=(41x-2x^2)/(1+.5pi) and take the derivative of that.

A'=(41-4x)/(1+.5pi). Set that equal to zero, which gives you x=41/4.

Make sure that is a max by taking the second derivative. A''<0 which implies a max at x=41/4

Put x=41/4 back into the d=(41-2x)/(1+.5pi) that we found earlier. Now you have x and d so find area and you are done.

Answer 3

Let the length of the rectangle be y ft and its width be 2x ft

Then Perimeter = 2y + 2x + ½π(2x)
= 2y + (π + 2)x
= 41

So y = ½( 41 - (π + 2)x)

Area = 2xy + ½πx²
= 2x * ½( 41 - (π + 2)x) + ½πx²

= 41x - (π + 2)x² + ½πx²

= 41x - ½(2π + 3)x²

dA/dx = 41 - (2π + 3)x
= 0 for stationary points (which is a maximum for this quadratic expression)

So at maximum area x = 41/(2π + 3) ft

Thus Amax = 41*41/(2π + 3) - ½(2π + 3) * [41/(2π + 3)]²

= ½(41/(2π + 3))²

≈ 9.7531 ft²