Posted by **Anonymous** on Saturday, November 5, 2011 at 4:07pm.

A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus the diameter of the semicircle is equal to the width of the rectangle.) If the perimeter of the window is 30ft, find the dimensions of the window so that the greatest possible amount of light is admitted.

What I have so far:

C = 1/2*pi*d, where d=x, or the width of the rectangle

P(rectangle) = 2y + x

So, the total perimeter:

P = 2y + x + 1/2*pi*x = 30ft

I'm not sure what to do next. How do I make this expression in terms of x?

Thank you

- Calculus -
**Damon**, Saturday, November 5, 2011 at 5:09pm
area = A = pi x^2/8 + x y

so

y = [A - pi x^2/8]/x

so

30 = 2 [A - pi x^2/8]/x + x + .5 pi x

30 x = 2 A - pi x^2/4 + x^2 + .5 pi x^2

(1+pi/4)x^2 - 30 x = -2 A

parabola, find vertex for max

1.785 x^2 - 30 x = -2 A

x^2 - 42 x = -1.12 A

complete square

x^2 -16.8 x + 70.6 = -1.12 A + 70.6

(x-8.4)^2= -1.12 A + 70.6

= -1.12 (A -63.3)

so x = 8.4 at vertex

and A = 63.3

so find y from

y = [A - pi x^2/8]/x

= [ 63.3 - pi 70.6/8] /8.4

=4.23

Check my arithmetic !

=

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