# Find max/min values.

• Nov 15th 2012, 07:42 PM
Find max/min values.
Find the max and min values of f(x,y)=x^2 + y^2 + 4 over the region R={(x,y) : x^2 + 2(y^2) =< 3
• Nov 15th 2012, 07:58 PM
MarkFL
Re: Find max/min values.
You want to equate the first partials to zero and solve the resulting system to find the critical points:

\$\displaystyle f_x(x,y)=0\$

\$\displaystyle f_y(x,y)=0\$

Solve the resulting system to find any critical points in the given region.

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

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

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

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

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

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

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