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Homework Help: Maximum and Minimum problem

  1. Jul 19, 2010 #1
    1. The problem statement, all variables and given/known data

    11. Maximize Q=xy, where x and y are positive numbers such that (4/3)x2 + y=16

    2. Relevant equations



    3. The attempt at a solution

    I know how to do it. The first time I did it. I multiplied the (4/3)x^2 +y=16 by (3/4) and got x^2+y=12 and then y=12-x^2 when solving for y. I used that equation to find the derivative of the product xy.

    How come I cannot do that because this still gave me a maximum and the critical point was still 2 but the other numbers were wrong. I don't understand why you can't solve for y like that?
     
  2. jcsd
  3. Jul 19, 2010 #2

    Mark44

    Staff: Mentor

    Maybe you did it correctly the first time, but what you have above is incorrect. If you multiply (4/3)x2 + y = 16 by 3/4, you get x2 + (3/4)y = 12.

    In any case, there is no need to multiply by 3/4. Just solve for y in that equation, and substitute for y in Q = xy.
    The two critical numbers are x = 2 and x = -2. Since x > 0, you don't need to worry about x = -2. Because of the error in forgetting to multiply y by 3/4, that affects the value you get for y, so I suspect that's why you're getting incorrect value for the maximum value.
     
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