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Applied maximum and minimum problems

  1. Oct 16, 2008 #1
    Optimization

    1. A new branch bank is to have a floor area of 3500 ft[tex]^{2}[/tex]. It is to be a rectangle w/ 3 solid brick walls and a decorative glass front. The glass costs 1.8 times as much as the brick wall per linear foot. What dimensions of the building will minimize cost of materials for the walls and front?

    I'm learning this stuff on my own and I need a little guidance on this problem. I know I have to write everything in terms of a single variable before I can derive. I'm having trouble understanding how to set up the problem. Am I looking to minimize surface area and does the volume come into play here? Thanks.

    So I have y=3500/x. Do I have to use that in surface area or volume?
     
    Last edited: Oct 16, 2008
  2. jcsd
  3. Oct 17, 2008 #2

    HallsofIvy

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    Staff Emeritus
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    Re: Optimization

    Neither, actually. Since you are not given any height you cannot find either the volume of the building or the area of the walls. And you do not want to miniize surface area: you are specifically asked to minimize the "cost".

    You are given the cost in terms of "linear feet" so the cost of the wall will be proportional to its length. If x is the length of the front and back and y is the length of the sides, then we have 2 brick walls of length y and one brick wall of length x. Take the "cost" of one linear foot of brick wall to be "1". Then the total cost of brick walls will be 2y+ x. The remaining wall, of length x, is glass and its cost will be 1.8x. The total cost of the walls is 2y+ x+ 1.8x= 2y+ 2.8x. That is what you want to minimize subject to the condition that xy= 3500. Yes, y= 3500/x and you can replace y by that in 2y+ 2.8x to get a function of x only.
     
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