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Maximization problem

Date: Tue, 29 Nov 1994 14:48:50 -0400 (EDT)
From: Anonymous
Organization: Wheelersburg HS

A window is to have the shape of a rectangle topped by a semi-circle.
Suppose that the semi-circle part of the window admits one-half as much
light per square foot as does the rectangular part. What are the dimensions
x and y of the window admitting the most light?

Date: 7 Dec 1994 02:39:54 GMT
Subject: Re: MATH STUFF


     Thanks for writing back and clarifying this problem.

     Well, if the total perimeter of the window is twenty then we have
something like this.

     Call the radius of the circle R and the length of the side of the
rectangle, y; then the base of the rectangle x is equal to 2r.

     So we have two equations.     

      Perimeter = 20 = 2y + 2r + Pi r  


      Light = 2yr + (Pi/4) r^2

Do you see how I got these equations? If not then write back and I will
explain it.

After you have these equations and understand them, then solve the first
one for y and you get:

            y = 10 - r - (Pi/2) r

Then you need to substitute into the equation for light.

            Light = 2( 10 - r - (Pi/2) r ) r  + (Pi/4) r^2

     Now we have reduced the problem to one equation based upon one variable
and we just have a normal maximization problem.  Do you know how to
proceed from here?

If not then write back and somebody will help you out.

      Ethan Doctor On Call
Associated Topics:
High School Euclidean/Plane Geometry
High School Geometry

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