# Finding maxima and minima of two variable function

• Dec 15th 2009, 08:52 AM
Dwill90
Finding maxima and minima of two variable function
Find the local maximum and minimum values and saddle points of the function :

f(x,y) = x^3 - 6xy + 8y^3

I found the critical points (0,0) and (1,1/2) by setting the partial derivatives equal to zero and solving for x and y.

Then, I tried the second derivative test:
D = fxx(a, b) * fyy (a, b) - [fxy(a, b)]^2
But found that D(0,0) = 0, meaning that the second derivative test fails for the point (0,0). What do I do when the second derivative test fails? Note that I am not allowed a graphing calculator.

Thank you,
David

• Dec 15th 2009, 09:56 AM
tonio
Quote:

Originally Posted by Dwill90
Find the local maximum and minimum values and saddle points of the function :

f(x,y) = x^3 - 6xy + 8y^3

I found the critical points (0,0) and (1,1/2) by setting the partial derivatives equal to zero and solving for x and y.

Then, I tried the second derivative test:
D = fxx(a, b) * fyy (a, b) - [fxy(a, b)]^2
But found that D(0,0) = 0, meaning that the second derivative test fails for the point (0,0). What do I do when the second derivative test fails? Note that I am not allowed a graphing calculator.

Thank you,
David

$f_x=3x^2-6y\,,\,f_{xx}=6x\,,\,\,f_y=-6x+24y^2\,,\,f_{yy}=48y,
f_{xy}=f_{yx}=-6\Longrightarrow$
$\left(f_{xx}f_{yy}(0,0)\right)-\left(f_{xy}(0,0)\right)^2=0-(-6)^2=-36$ ...why do you say it is zero?

Tonio
• Dec 15th 2009, 10:15 AM
Dwill90
I see, I miscalculated the partial derivative fxy, thinking it to be zero instead of -6. Thanks for the help!