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sagar12345 · Oct 29, 2009, 07:07 AM

Find the critical points of the function f(x,y)=x^3+y^2-12x-6y+40 test each of these for maximum and minimum

sagar12345 · Oct 29, 2009, 07:10 AM

Find the critical points of the function f(x,y)=x^3+y^2-12x-6y+40 test each of these for maximum and minimum

sagar12345 · Oct 29, 2009, 07:11 AM

Find the critical points of the function f(x,y)=x^3+y^2-12x-6y+40 test each of these for maximum and minimum

galactus · Oct 29, 2009, 08:17 AM

Please. You do not have to post it three times.

Anyway, are you famiiar with partial differenitation?

$f(x,y)=x^{3}+y^{2}-12x-6y+40$

Find the derivative w.r.t x treating y as a constant:

$f_{x}(x,y)=3x^{2}-12$

Derivative w.r.t y treating x as a constant:

$f_{y}(x,y)=2y-6$

The critical points satisfy:

$3x^{2}-12=0$ and

$2y-6=0$

Solve for x and y.

Then use the Second Partials Test to see if the critical point is max, min, or saddle point.

$D=f_{xx}(x,y)f_{yy}(x,y)-f_{xy}^{2}(x,y)$

If D>0 and $f_{xx}>0$ , then we have a relative minimum

If D>0 and $f_{xx}<0$ , then we have a relative maximum.

If $D<0$ , then there is a saddle point.

If D=0, then no conclusion can be drawn.

Are you familiar with test? It should be in your calc book.

Another course may be to perform implicit differentiation. Then, look at what makes the numerator 0 and what makes the denominator 0.

$y'=\frac{-3(x^{2}-4)}{2(y-3)}$

See if they are the same as with the other method. I bet they are.