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anonymous

  • 5 years ago

Maximize Q = xy2, where x and y are positive numbers, such that x + y2 = 4.

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  1. anonymous
    • 5 years ago
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    does y2 means y squared?

  2. anonymous
    • 5 years ago
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    yes

  3. anonymous
    • 5 years ago
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    If so then the solution of the problem is: You want to maximize Q, that means you need to find the derivative and critical points of Q. The problem here is that Q is in terms of x and y. You should make it in term of only one of them. So we are going to use the second equation to do so. \[x+y^2=4 \implies y^2=4-x\] substitute for the value of y^2 into Q, you get: \[Q=xy^2=x(4-x)=4x-x^2\]

  4. anonymous
    • 5 years ago
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    Do you know how to maximize 4x-x^2?

  5. anonymous
    • 5 years ago
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    I plug test points into 4x-x^2?

  6. anonymous
    • 5 years ago
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    the critical number is 4?

  7. anonymous
    • 5 years ago
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    ^^ yeah exactly.

  8. anonymous
    • 5 years ago
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    so the max is 3?

  9. anonymous
    • 5 years ago
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    you solved for x when the derivative is zero, and you got tgat x=4. this is your only critical point, so it's either maximum or minimum. Which one do you think it is?

  10. anonymous
    • 5 years ago
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    oooh. i dont know

  11. anonymous
    • 5 years ago
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    hey wait!!

  12. anonymous
    • 5 years ago
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    I made a mistake :(

  13. anonymous
    • 5 years ago
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    lol oh no

  14. anonymous
    • 5 years ago
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    the critical point is not x=4

  15. anonymous
    • 5 years ago
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    I didn't make it, you did and I didn't see it :P

  16. anonymous
    • 5 years ago
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    Don't worry.. It's so easy. Do you know how to find the derivative of 4x-x^2?

  17. anonymous
    • 5 years ago
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    4- 2x

  18. anonymous
    • 5 years ago
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    Right!! Now solve for x when 4-2x=0.

  19. anonymous
    • 5 years ago
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    2!

  20. anonymous
    • 5 years ago
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    haha so hard

  21. anonymous
    • 5 years ago
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    LOL

  22. anonymous
    • 5 years ago
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    Now 2 is your critical point, and at x=2 Q is a maximum value.

  23. anonymous
    • 5 years ago
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    So now just plug x=2 in Q=4x-x^2, and you're done!!

  24. anonymous
    • 5 years ago
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    yay thanks

  25. anonymous
    • 5 years ago
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    What is the maximum value?

  26. anonymous
    • 5 years ago
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    4

  27. anonymous
    • 5 years ago
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    :) I hope you can do any similar problem next time!

  28. anonymous
    • 5 years ago
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    thanks for your help!!

  29. anonymous
    • 5 years ago
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    no problem.

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