Local Extrema help

**square** · November 25th 2009, 05:43 AM

How do i find the "local extrema" for a function like this: f(x,y) = xy+y-15x

**Soroban** · November 25th 2009, 06:04 AM

Hello, square!

How do i find the "local extrema" for a function like this: . $f(x,y) \:=\: xy+y-15x$

Solve the system: . $\begin{array}{ccc}\dfrac{\partial f}{\partial x} &=& 0 \\ \\[-3mm] \dfrac{\partial f}{\partial y} &=& 0 \end{array}$

And test the critical value(s) in: . $D \;=\;\left(\frac{\partial^2\!f}{\partial x^2}\right)\left(\frac{\partial^2\!f}{\partial y^2}\right) - \left(\frac{\partial^2\!f}{\partial x\partial y}\right)^2$

If all this is meaningless to you,
. . (1) you should not have been assigned this problem, or
. . (2) you weren't paying attention in class.

**tonio** · November 25th 2009, 06:15 AM

Originally Posted by square

How do i find the "local extrema" for a function like this: f(x,y) = xy+y-15x

Solve the equations $f_x=y-15=0\,,\,\,f_y=x+1=0$ , and then use the Hessian's determinant in the points $(x,y)$ that you found above, $f_{xx}f_{yy}-(f_{xy})^2$ : if this determinant is positive and (1) $f_{xx}>0$ , then the point you found is a local minimum, and if (2) $f_{xx}<0$ the point is then a local maximum .

If the determinant is negative then the point is a saddle point (not max. not min.), and if the determinant is zero then the Hessian matrix's test fails to decide.

Tonio

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