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Thread: Is my set up for Optimization problem correct? very quick!

  1. #1
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    Is my set up for Optimization problem correct? very quick!

    The problem is below:
    Is my set up for Optimization problem correct? very quick!-capture.png
    Let x = the length of the edge of one cut out corner .

    So the objective function is : V = x(10-2x)^2 + x^3

    and there wouldn't be a constraint function other than that x >=0 and <= 5

    Is this the correct function that i would take derivative and find critical points for?
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    Re: Is my set up for Optimization problem correct? very quick!

    Quote Originally Posted by lc99 View Post
    The problem is below:
    Click image for larger version. 

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    Let x = the length of the edge of one cut out corner .

    So the objective function is : V = x(10-2x)^2 + x^3

    and there wouldn't be a constraint function other than that x >=0 and <= 5

    Is this the correct function that i would take derivative and find critical points for?
    yes, however the constraint is $0 < x < 5$ ... no equality at the endpoints
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    Re: Is my set up for Optimization problem correct? very quick!

    Sometimes I'm confused cause my professor includes the endpoints so that we could apply the extreme value theory. Cause if we know that end points are 0, the must critical points
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    Re: Is my set up for Optimization problem correct? very quick!

    on 2nd thought, the usable domain for x would be $0 < x \le 5$

    $x = 0$ ... no boxes, therefore zero volume

    $x = 5 \implies V = 125$
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    Re: Is my set up for Optimization problem correct? very quick!

    Wait, how can x be just 5? if it was justs 5, then we dont cut anything??

    I'm not sure if my answer is correct if i do it like that...

    What i did was:

    V = x(10-2x^^2 + x^3
    = 5x^3 - 40x^2 +100x

    V' = 5(3x-10)(x-20 =0

    x = 10/3 , 2

    end points are x = 0 , x=5

    when x = 0, V = 0
    when x =2, V =80
    when x=10/3 , V= 2000/27
    when x = 5, V = 125

    The maximum volume is when x =5 , but doesn't x = 5 mean that there were no squares cut, therefore, you cant really form a paper clip fold with top and bottom not closed? Doesn't the problem ask that there should be both items involved? or not?
    Last edited by lc99; Nov 9th 2017 at 11:42 AM.
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    Re: Is my set up for Optimization problem correct? very quick!

    Quote Originally Posted by lc99 View Post
    Wait, how can x be just 5? if it was justs 5, then we dont cut anything??
    yes you do ... four 5x5 pieces to make the paper clip "box"
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