global maxima and minima: f(x, y) = x^2 + y^2 - 2x

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Erin0702

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Find the global maximum value and global minimum value of f(x,y) on region S and state at which points they occur.

f(x,y)=x²+y²-2x

S is the triangular region (including the boundary) with vertices at (0,0), (2,0), and (0,2).

In previous problems they represented the boundary parametrically with cos and sin being substituted for x and y. I'm not too sure where to go with this one since I have been given vertices.

Any help is greatly appreciated!
 
You need to use the derivatives and also check the boundary points.
 
but if I use the derivatives and just check the boundary points isn't it only going to give me local minima and maxima? I thought I had to do something with parametric equations for global extrema??
 
The region S is the triangle with equation \(\displaystyle y=-x+2\)

Partial derivatives of \(\displaystyle x^{2}+y^{2}-2x\)........[1]

\(\displaystyle \frac{{\partial}f}{{\partial}x}=2x-2\)

\(\displaystyle \frac{{\partial}f}{{\partial}y}=2y\)

Solve for x and y, respectively:

\(\displaystyle 2x-2=0\), x=1

\(\displaystyle 2y=0\), y=0

(1,0) is the only crirical point.

You have 3 line segments. Treat each separately.

There should be an example of this sort of problem in your calc text, maybe under maxima and minima of two variables.
 
In my book they have an example of a triangle where you find the maximum and minimum values. Following this example, I need to find the equation of each side of the triangle and obtain x and y values.

Side 1 (from vertices (0,0) to (2,0))
slope =0
so the equation is y=0
z(x,y)=x^2+y^2-2x
plugging in 0 for the x the equation becomes:
z(x,y)=x^2-2x
z'(x,y)=2x-2
2x=2
x=1
the point becomes (1,0) from plugging it into y=0

Side 2 (from vertices (2,0) to (0,2))
slope = -1
y-int = 2
equation=y=-x+2
plug the y into z(x,y)
z(x,y)=x^2+(-x+2)^2-2x
=2x^2-6x+4
2(x-1)(x-2)
so x=1 and x=2
the points are (1,1) and (2,0)

Side 3 (from vertices (0,0) to (0,2))
slope=0/0 which is undefined

This is where I got confused...does it mean I don't have any points from this line?

And then from here the book made a table with all of the x,y,z values obtained from this method and I would include the critical point here. Then they found the minimum and maximum value.

Is this the same method I should follow or have I messed something up?

thanks!
 
Erin0702 said:
but if I use the derivatives and just check the boundary points isn't it only going to give me local minima and maxima? I thought I had to do something with parametric equations for global extrema??
Click to expand...
This is a great moment in the life of all mathematics students. It is the moment when you realize that mathematics never was and never will be a dry or arid science. You must use judgment, skill, and experience in order to determine the difference between local and global. It isn't always obvious. In the Real World, it's usually less obvious.

Welcome!

P.S. With apologies to Sofia Kovalevskaya.
 
Erin0702 said:
In my book they have an example of a triangle where you find the maximum and minimum values. Following this example, I need to find the equation of each side of the triangle and obtain x and y values.

Side 1 (from vertices (0,0) to (2,0))
slope =0
so the equation is y=0
z(x,y)=x^2+y^2-2x
plugging in 0 for the x the equation becomes:
z(x,y)=x^2-2x
z'(x,y)=2x-2
2x=2
x=1
the point becomes (1,0) from plugging it into y=0

Side 2 (from vertices (2,0) to (0,2))
slope = -1
y-int = 2
equation=y=-x+2
plug the y into z(x,y)
z(x,y)=x^2+(-x+2)^2-2x
=2x^2-6x+4
2(x-1)(x-2)
so x=1 and x=2
the points are (1,1) and (2,0)

\(\displaystyle w(x)=2x^{2}-6x+4\), \(\displaystyle w'(x}=4x-6\)


Side 3 (from vertices (0,0) to (0,2))
slope=0/0 which is undefined

\(\displaystyle \text{For x=0, we have}\,\ v(y)=y^{2}\,\ v'(y)=2y\)

\(\displaystyle \text{There's a difference between local\\ and absolute extrema. Locate all your critical points(local extrema) and\\ check them in the original function. The one that gives you the lowest\\ value is the absolute mini, the one that gives you the largest value is the\\ absolute maxi.\\
As the book done}\)



This is where I got confused...does it mean I don't have any points from this line?

And then from here the book made a table with all of the x,y,z values obtained from this method and I would include the critical point here. Then they found the minimum and maximum value.

Is this the same method I should follow or have I messed something up?

thanks!
Click to expand...
 
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