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Thread: maximum and minimum

  1. #1
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    maximum and minimum

    given $\displaystyle g(x,y)=ax + by^2$


    $\displaystyle a>0,b>0,x+y=1, 0<=x>=1, 0<=y>=1$

    $\displaystyle a=1$ and $\displaystyle b=5/6$

    What is the smallest value y can have? can anyone show how to solve?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by EricTu View Post
    given $\displaystyle g(x,y)=ax + by^2$


    $\displaystyle a>0,b>0,x+y=1, 0<=x>=1, 0<=y>=1$

    $\displaystyle a=1$ and $\displaystyle b=5/6$

    What is the smallest value y can have? can anyone show how to solve?
    TYpe this again exactly as question was asked please.

    CB
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  3. #3
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    I assume you are asking for the minimum value $\displaystyle ax+ by^2$ on a regions bounded by some lines buy $\displaystyle 0<= x>= 1$ and $\displaystyle 0<=y>=1$ make no sense. If $\displaystyle x>= 0$ then x must be "$\displaystyle >= 0$. Did you mean "$\displaystyle 0<= x< = 1$ and $\displaystyle 0<= y<= 1$?

    If so then the minimum value must occur:
    1) in the interior of the set where the gradient is 0 (or undefined) or
    2) on the boundary of the set.

    $\displaystyle \nabla ax+ by^2= a\vec{i}+ 2by\vec{j}$ which is never 0 because of the "$\displaystyle a\vec{i}$" term. On the line x= 0, the y axis, the function is $\displaystyle by^2$ which has its minimum at y= 0. On the line y= 0, the x-axis, the function is $\displaystyle ax$ which has its minimum at x= 0. On the line x+ y= 1, x= 1- y so the function is $\displaystyle a- ay+ by^2$. The derivative of that is $\displaystyle -a+ 2by$ and is 0 at $\displaystyle y=\frac{a}{2b}$. Then $\displaystyle x= 1- \frac{a}{2b}= \frac{2b- a}{2b}$.

    Of course, we also need to consider the vertices (1, 0) and (0, 1). That is, the minimum value must occur at one of (0, 0), (1, 0), (0, 1), or $\displaystyle \left(\frac{2b-a}{2b}, \frac{a}{2b}\right)$. Evaluate $\displaystyle ax+ by^2$ at each of those points to decide where it is smallest.
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    I assume you are asking for the minimum value $\displaystyle ax+ by^2$ on a regions bounded by some lines buy $\displaystyle 0<= x>= 1$ and $\displaystyle 0<=y>=1$ make no sense. If $\displaystyle x>= 0$ then x must be "$\displaystyle >= 0$. Did you mean "$\displaystyle 0<= x< = 1$ and $\displaystyle 0<= y<= 1$?

    If so then the minimum value must occur:
    1) in the interior of the set where the gradient is 0 (or undefined) or
    2) on the boundary of the set.

    $\displaystyle \nabla ax+ by^2= a\vec{i}+ 2by\vec{j}$ which is never 0 because of the "$\displaystyle a\vec{i}$" term. On the line x= 0, the y axis, the function is $\displaystyle by^2$ which has its minimum at y= 0. On the line y= 0, the x-axis, the function is $\displaystyle ax$ which has its minimum at x= 0. On the line x+ y= 1, x= 1- y so the function is $\displaystyle a- ay+ by^2$. The derivative of that is $\displaystyle -a+ 2by$ and is 0 at $\displaystyle y=\frac{a}{2b}$. Then $\displaystyle x= 1- \frac{a}{2b}= \frac{2b- a}{2b}$.

    Of course, we also need to consider the vertices (1, 0) and (0, 1). That is, the minimum value must occur at one of (0, 0), (1, 0), (0, 1), or $\displaystyle \left(\frac{2b-a}{2b}, \frac{a}{2b}\right)$. Evaluate $\displaystyle ax+ by^2$ at each of those points to decide where it is smallest.
    Yes, sorry - it was a typo. The way you write it is correct.
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  5. #5
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    The question is: Assume that a=1 and b=5/6. What is the minimum value y can have in given boundaries (0<=y<=1)?
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