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Optimization, multivariable calculus

  1. Apr 23, 2012 #1
    1. The problem statement, all variables and given/known data
    the base of an aquarium with given volume V is made of slate and the sides are made of glass. if slate costs five times as much (per unit area) as glass, find the dimensions of the aquarium that minimize the cost of the materials.


    2. Relevant equations



    3. The attempt at a solution
    so this problem fall under finding max, min and saddle points of a function
    i think that lagrange multipliers can also be used but id like to first solve it by finding critical points and then the minimum
    so first i know that V=xyz and and the total area of the glass will be 2xz+2yz while the area of the base slate will be xy
    then the cost function will be C(x,y,z)=the cost of the slate + the cost of the glass
    so then C(x,y,z)=5(2xz+2yz)+(2xz+2yz)= 12xz+12yz=z(12x+12y)(?)
    im not really sure what to do next, but what i though of doing was to solve for z in V=xyz, then using that z by replacing it for the z in C(x,y,z) then i would only have a function of 2 variables C(x,y) and then i can just go ahead and find partial derivatives and use the second derivative test to find the minimum
    is this the correct way to go about finding the dimensions?
     
  2. jcsd
  3. Apr 23, 2012 #2

    Ray Vickson

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    Try it and see! The only real way to learn is to go ahead and do it. (Note: an alternative would be to use a Lagrange multiplier method.)

    RGV
     
  4. Apr 23, 2012 #3
    i actually did try right at after posting my question
    heres what i got
    C(x,y,z)=z(12x+12y)
    solving for z in V=xyz gives me z=V/(xy)
    then C(x,y)=V/(xy)(12x+12y)=12V/y+12V/x
    C_x(partial derivative)=-12V/x^2 and C_y=-12V/y^2
    this leads to a problem because then i have no critical points
    any suggestions?
     
  5. Apr 23, 2012 #4

    Ray Vickson

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    You made a error writing down the cost. Cost = 2xz+2yz + 5xy [=cost of 2 sides + cost of 2 ends + cost of bottom---assuming no top]. When you use the volume constraint to eliminate z, you get an expression C(x,y) that does have a stationary point in x>0,y>0.

    RGV
     
  6. Apr 23, 2012 #5
    ohh sweet thanks!
    i left a question mark after writing down the cost function because i wasnt quite sure i had the right function for the cost
     
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