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Math Help - maximum and minimum

  1. #1
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    maximum and minimum

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


    a>0,b>0,x+y=1, 0<=x>=1, 0<=y>=1

    a=1 and 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 g(x,y)=ax + by^2


    a>0,b>0,x+y=1, 0<=x>=1, 0<=y>=1

    a=1 and 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 ax+ by^2 on a regions bounded by some lines buy 0<= x>= 1 and 0<=y>=1 make no sense. If x>= 0 then x must be " >= 0. Did you mean " 0<= x< = 1 and 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.

    \nabla ax+ by^2= a\vec{i}+ 2by\vec{j} which is never 0 because of the " a\vec{i}" term. On the line x= 0, the y axis, the function is by^2 which has its minimum at y= 0. On the line y= 0, the x-axis, the function is ax which has its minimum at x= 0. On the line x+ y= 1, x= 1- y so the function is a- ay+ by^2. The derivative of that is -a+ 2by and is 0 at y=\frac{a}{2b}. Then 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 \left(\frac{2b-a}{2b}, \frac{a}{2b}\right). Evaluate 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 ax+ by^2 on a regions bounded by some lines buy 0<= x>= 1 and 0<=y>=1 make no sense. If x>= 0 then x must be " >= 0. Did you mean " 0<= x< = 1 and 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.

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