# Extreme Value Theorem

• Feb 11th 2009, 11:32 PM
toop
Extreme Value Theorem
Let S be the set of all points inside of the triangle whose vertices are

$(-2,0), (0,2), (2,0).$

Find the minimum and maximum values of the function

$f(x,y) = x^2 + 3y^2 - 4xy + x - y +1$

The equations for the sides of this triangle are:

$y=0 , -2 \leq x \leq 2$
$y = 2-x , 0 \leq x \leq 2$
$y = 2+x , -2 \leq x \leq 0$

From what I comprehended in class and in my notes, I substitute $y$ into the original function and then take the derivative of the result. I then set the derivative equal to 0 to find the critical points?

From there I am not too sure what to do...I assume I do it for all three equations and then plug it back into the original function to find the min/max?
• Feb 12th 2009, 12:41 AM
NonCommAlg
Quote:

Originally Posted by toop
Let S be the set of all points inside of the triangle whose vertices are

$(-2,0), (0,2), (2,0).$

Find the minimum and maximum values of the function

$f(x,y) = x^2 + 3y^2 - 4xy + x - y +1$

The equations for the sides of this triangle are:

$y=0 , -2 \leq x \leq 2$
$y = 2-x , 0 \leq x \leq 2$
$y = 2+x , -2 \leq x \leq 0$

From what I comprehended in class and in my notes, I substitute $y$ into the original function and then take the derivative of the result. I then set the derivative equal to 0 to find the critical points?

From there I am not too sure what to do...I assume I do it for all three equations and then plug it back into the original function to find the min/max?

this is a quite long exercise! first solve $f_x=f_y=0$ and choose those solutions that are inside your region. next consider the 3 sides of the triangle and find the values of $f$ in critical points and

end points of each side. finally points with largest or smallest value of $f$ give you the maximum and minimum of $f.$ to give you a better idea, i'll show you how to do it for one of the sides:

for example on $AB,$ where $A(0,2), \ B(2,0).$ the equation of the segment AB is: $y=2-x, \ 0 \leq x \leq 2.$ therefore on AB we have:

$f=x^2+3(2-x)^2-4x(2-x)+x - (2-x)+1=8x^2-18x+11, \ 0 \leq x \leq 2.$ now $f'=16x-18=0$ gives us: $x=\frac{9}{8},$ and thus $\boxed{f=\frac{7}{8}}$. you also need to check the end points, i.e. x = 0 and

x = 2. at x = 0, we have $\boxed{f=11}$ and at x = 2 we have $\boxed{f=7}$. now do the same for other two sides.
• Feb 12th 2009, 11:29 AM
toop
I appreciate the help!

When solving for $f_x = f_y = 0$ I came up with:

$f_x = 2x-4y+1$
$f_y = 6y-4x-1$

From there, I am not sure how to set it to 0?

I tried $x = (4y-1) / 2$ but not sure where to go from there.

Also, I understood how you came up with $f = 7/8 , f = 11, f = 7$ .... do I then just plug it into the original function?

so: $f(9/8,7/8), f(0,11)$ and so on?
• Feb 12th 2009, 09:29 PM
NonCommAlg
Quote:

Originally Posted by toop
I appreciate the help!

When solving for $f_x = f_y = 0$ I came up with:

$f_x = 2x-4y+1$
$f_y = 6y-4x-1$

From there, I am not sure how to set it to 0?

I tried $x = (4y-1) / 2$ but not sure where to go from there.

Also, I understood how you came up with $f = 7/8 , f = 11, f = 7$ .... do I then just plug it into the original function?

so: $f(9/8,7/8), f(0,11)$ and so on?

no! for $x=\frac{9}{8}$ we get $y=\frac{7}{8}$ and $f(\frac{9}{8}, \frac{7}{8})=\frac{7}{8}.$ similarly for x = 0 and x = 2 we have y = 2 and y = 0 respectively and: $f(0,2)=11, \ f(2,0)=7.$ in order to find the value of $f$ you don't need to

find the value of y because we already have $f$ as a function of $x$ only. so you only need $x.$ that's why i didn't mention y in my previous post.
• Feb 12th 2009, 11:05 PM
toop
After all is said and done...can you verify my answers:

Min = -5

Woops, I meant MAX = 11
• Feb 12th 2009, 11:25 PM
NonCommAlg
Quote:

Originally Posted by toop
After all is said and done...can you verify my answers:

Max = 7

Min = -5

that's not correct! according to my calculations, which are almost always correct! (Evilgrin), the maximum of $f$ is $11$ and the minimum is $\frac{3}{4}.$
• Feb 12th 2009, 11:28 PM
toop
Ah, stupid little math errors I made.

I now have the same answers as you of min=.75 and max=11.

Thank you very much!