# maximum and minimum

• Dec 15th 2010, 05:31 AM
EricTu
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?
• Dec 15th 2010, 05:36 AM
CaptainBlack
Quote:

Originally Posted by EricTu
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?

CB
• Dec 15th 2010, 05:44 AM
HallsofIvy
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.
• Dec 15th 2010, 05:52 AM
EricTu
Quote:

Originally Posted by HallsofIvy
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.
• Dec 15th 2010, 05:59 AM
EricTu
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)?