# need help

• November 14th 2012, 06:48 AM
Petrus
need help
Determine the maximum and minimum of the function
https://webwork.math.su.se/webwork2_...9ef36cb0b1.png
in the square having corners at points https://webwork.math.su.se/webwork2_...3eb6251e51.png and https://webwork.math.su.se/webwork2_...9aaad4a871.png.
my progress:

derivate on x i get $12x^2-6y$
derivate on y i get $12y-6x$
idk if this is right thinking but i wanna solve x and i get
$y=6x/12$
so i got $12x^2-6x/12=0$ and we get that x_1=0 and x_2=1/24 i am right or wrong and what shall i do next?

• November 14th 2012, 12:08 PM
MarkFL
Re: need help
You have the right idea. You want to equate the partials to zero and solve the resulting system to find the critical points:

$f_x(x,y)=12x^2-6y=0$

$f_y(x,y)=12y-6x=0$

Solve this system, we find:

$(x,y)=(0,0),\,\left(\frac{1}{4},\frac{1}{8} \right)$

Now you want to use the second partials test for relative extrema:

Let $(a,b)$ be a critical point of $z=f(x,y)$ and suppose $f_{xx},\,f_{yy}$ and $f_{xy}$ are continuous in a rectangular region containing $(a,b)$.

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

i) If $D(a,b)>0$ and $f_{xx}(a,b)>0$, then $f(a,b)$ is a relative minimum.

ii) If $D(a,b)>0$ and $f_{xx}(a,b)<0$, then $f(a,b)$ is a relative maximum.

iii) If $D(a,b)<0$, then $f(a,b)$ is not an extremum.

iv) If $D(a,b)=0$, then no conclusion can be drawn concerning a relative extremum.

Using this test, what conclusions can be drawn about the critical points?