Maximization problemDate: Tue, 29 Nov 1994 14:48:50 -0400 (EDT) From: Anonymous Subject: MATH STUFF 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 Hey, 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 And 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 |
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