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Calculus 3 - Absolute Min

  1. Nov 5, 2006 #1
    Here is the question:

    A closed rectangular box with a volume of 16ft^3 is made from two kinds of materials. The top and bottom are made of material costing 10 cents per square foot and the sides from material costing 5 cents per square foot. Find the dimensions of the box so that the cost of materials is minimized.

    Here is how I've been attempting to do it (but can't get the answer to match the one given in the back of the book):

    The volume of the box is the constraint and that formula is V = xyz = 16
    The cost of the box is the function given by .20xy + .10zy + .10zx

    Okay... so I solved the constraint for one variable and got z=(16)/(xy) and then I substituted this into the cost equation of the box for z. And this gave me 2xy + 16/x + 16/y.

    Solving for the partial derivatives, I got f(x) = 2y - 16(x^-2) and f(y) = 2x - 16(y^-2). I set these equal to 0. Then I took the f(x) and solved for y to get y = 8 (1/(8x^-2)^2). Then I plugged this into the f(y) forumal to solve for x and got x = 8, so y = 1/8. So then the critical point from all of this is (8, (1/8)).

    Then since it is looking for the absolute minimum, I set up up the problem to find the critical points on the boundaries of the box in the xy plane. I got (0,16), (16,0), (16,1), (1,16) as critical points, along with the (8,(1/8)) from above and the end points (0,0) and (16,16).

    I know I'm messing up somewhere and I'm not entirely sure where to go from here. The answer the book gives is x=2, y=2, and z=4.

    Can anyone help me out here? The more I try to redo it, the more I get confused....
     
  2. jcsd
  3. Nov 6, 2006 #2

    HallsofIvy

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    No. From 2y- 16x-2= 0 you get y= 8x-2! I think you must have put that into fy and forgot you had done that!

    I wouldn't have done it quite that way. From 2y-15x-2= 0 and 2x- 16y-2= 0 you can get 2x2y= 16 and 2xy2= 16. Dividing one equation by the other, x/y= 1 or x= y.(which was obvious from the first!) Now, each equation is 2x3= 16 or x3= 8 so x= 2, not 8. Did you lose a 3?
    From x= 2, y= 2 and then, to have xyz= 16, z= 4.


    No, those don't satisfy xyz= 16. The possible values of x,y,z approach those but can't equal them. Since this is an open set, there is not a maximum. The minimum cost is when x= y= 2, z= 4.

     
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