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Local extrema w/2 var

  • Thread starter Guero
  • Start date
15
0
we're doing partial derivatives, and i thought i understood them until:

[tex]f(x,y)=-x^2-y^2-10xy+4y-4x+2 [/tex]

we are meant to find and classify all local extrema. i got:

[tex]f_x=-2x-10y-4 [/tex] and [tex] f_y=-2y-10x+4[/tex]

i've never had two variables at this stage, and i can't solve for any of the points. the text doesn't seem to say anything about this...
 

Answers and Replies

15
0
yay! i'm now a level 30 thumb twiddler
 
TD
Homework Helper
1,020
0
You solve [tex]\nabla f = \vec0[/tex] (gradient of f) which basically gives you a system of 2 equations here. The solutions are the 'stationary points' which may be extrema, but you have to check that (e.g. with a test)

So you solve:

[tex]\left\{ \begin{gathered}
\frac{{\delta f}}
{{\delta x}} = 0 \hfill \\
\frac{{\delta f}}
{{\delta y}} = 0 \hfill \\
\end{gathered} \right.[/tex]
 
15
0
mm, i get: [tex]10y=-2x-4 [/tex] and [tex]10x=-2y+4 [/tex] for each of which there are infinite solutions for x and y. i thought that if i combined the equations (no reason) it might do it, but i still got [tex]y=x-1 [/tex]
 
TD
Homework Helper
1,020
0
Yes, you have to combine them in a system.

[tex]\left\{ \begin{gathered}
2x + 10y + 4 = 0 \hfill \\
10x + 2y - 4 = 0 \hfill \\
\end{gathered} \right.[/tex]

Solving it should give [itex]\left( {\frac{1}
{2}, - \frac{1}
{2}} \right)[/itex]

However, although it's a stationary point, it's not an extremum.
 
15
0
aha! thanks, i get it now
 

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