# Extrema points

• January 13th 2009, 12:34 AM
tsal15
Extrema points
"find and classify the extreme point $(x,y)$ of the function of two variables $f(x,y) = x^2 + y^2 - xy$" States the question

My solution:

dz/dx = 2x - y

dz/dy = 2y - x

y = x = 0

therefore the extrema point is at (0,0)

to classify it:

$D = \frac{d^2z}{dx^2} . \frac{d^2z}{dy^2} - \frac{d^2z}{dxdy}
$

$D = 2 . 2 - (-1)^2 = 3$

3 > 0 which means the extrema is a local minima

Am I correct?

Thank you :)
• January 13th 2009, 12:57 AM
earboth
Quote:

Originally Posted by tsal15
"find and classify the extreme point $(x,y)$ of the function of two variables $f(x,y) = x^2 + y^2 - xy$" States the question

My solution:

dz/dx = 2x - y

dz/dy = 2y - x

y = x = 0

therefore the extrema point is at (0,0,0)

to classify it:

$D = \frac{d^2z}{dx^2} . \frac{d^2z}{dy^2} - \frac{d^2z}{dxdy}
$

$D = 2 . 2 - (-1)^2 = 3$

3 > 0 which means the extrema is a local minima

Am I correct?

Thank you :)

Yes. The graph of your function is a paraboloid with its vertex at (0, 0, 0)
• January 13th 2009, 04:41 AM
tsal15
Quote:

Originally Posted by earboth
Yes. The graph of your function is a paraboloid with its vertex at (0, 0, 0)

Why is it (0, 0, 0) ?

Thanks for your previous and future help :)
• January 13th 2009, 05:51 AM
earboth
Quote:

Originally Posted by tsal15
"find and classify the extreme point $(x,y)$ of the function of two variables $f(x,y) = x^2 + y^2 - xy$" ...

Quote:

Originally Posted by tsal15
Why is it (0, 0, 0) ?

Thanks for your previous and future help :)

As you've stated you have a function of two (independent) variables. That means the equation

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

describes:

$f:~\begin{array}{lcl}\mathbb{R} \times \mathbb{R}&\; \mapsto \ & \mathbb{R} \\ (x,y) &\ \mapsto\ & z=f(x,y)\end{array}$

Therefore the graph of this function consists of all points (x, y, z) whose coordinates satisfy the given conditions.
• January 13th 2009, 06:03 AM
tsal15
Quote:

Originally Posted by earboth
As you've stated you have a function of two (independent) variables. That means the equation

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

describes:

$f:~\begin{array}{lcl}\mathbb{R} \times \mathbb{R}&\; \mapsto \ & \mathbb{R} \\ (x,y) &\ \mapsto\ & z=f(x,y)\end{array}$

Therefore the graph of this function consists of all points (x, y, z) whose coordinates satisfy the given conditions.

Of course!!! great Thank YOU :)