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Thread: Minimizing 3D function

  1. #1
    Super Member sakonpure6's Avatar
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    Minimizing 3D function

    How would you minimize the following function with the constraints:

    $f(x,y) = \frac{75}{xy^2} ; x>=100, y>=200 , x \cdot y >=1.2 (100*200)$

    Can some one please provide me with some resource to help me minimize this? or help walk me through it?

    Thank you,
    Last edited by skeeter; Dec 14th 2017 at 10:47 AM. Reason: edited inop tex tags
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  2. #2
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    Re: Minimizing 3D function

    Quote Originally Posted by sakonpure6 View Post
    How would you minimize the following function with the constraints:

    $f(x,y) = \frac{75}{xy^2} ; x>=100, y>=200 , x\cdoty >=1.2 (100*200)$

    Can some one please provide me with some resource to help me minimize this? or help walk me through it?

    Thank you,
    Are you sure the first conditions are not $x\le 100$ and $y\le 200$? The constraints as given make this a rather strange figure!

    Any max or min that occurs in the interior of the figure must happen where the partial derivatives are 0. The partial derivative with respect to x is $-\frac{-75}{x^2y^2}$ and the derivative with respect to y is $-\frac{150}{xy^{-3}}$ which are never 0. Any max or min must be at a boundary.

    Assuming the region is as given in your post, the boundary has four parts, x= 10 for $0.12\le y\le 20$, y= 20 for $0.6\le x\le 10$, $y= \frac{1.2}{x}$ for $x< 0.6$, and $x= \frac{1.2}{y}$ for $y< 0.12$.

    On the first part of the boundary, x= 10, $f(10, y)= \frac{7.5}{y^2}$. Any min or max on that must be where the derivative is 0. But the derivative is $\frac{-15}{y^3}$ and, again, this is never 0. On the second part of the boundary, y= 20, $f(x, 20)= \frac{0.1875}{y^2}$. The derivative is $-\frac{0.375}{y^3}$ which is never 0. On the third part of the boundary, $y= \frac{1.2}{x}$, $f(x,y)= 52x$ so the derivative is the constant $52$ which again is never 0. Finally on the fourth part of the boundary, $y= \frac{0.06}{x}$ so that $f(x, y)= 20.83x$ so the derivative is the constant $208.3$, not 0.

    So any max or min must occur on the "boundary of the boundary" the points where those parts join. They are (10, 0.12), (0.06, 20), and (10, 20). Evaluate $f(x, y)= \frac{75}{xy^2}$ at each of those points to determine what the max and min are.
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  3. #3
    Super Member sakonpure6's Avatar
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    Re: Minimizing 3D function

    The conditions on the variables are as in my original post, i.e greater than. The above equation simply relates the maximum stress of a rectangular beam section ( f (x,y) ) in terms of its depth y and base x.
    The original problem was to minimize stress if we are allowed to increase cross sectional area by 20% ( thus the boundary condition x*y >= 1.2 (100*200) , original base x=100 and depth = 200)

    I am having a hard time following what you did with the original equations, what did you divide everything by?
    Can you also please elaborate a bit on how you obtained the boundry conditions, did you substitute x = 10 in the boundary condition x*y >= 1.2 (100*200) and then solve for the upper limit on y? How did you get the lower limit?

    Thank you,
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