# Local Max/Min Values using Partial Derivatives

• May 18th 2009, 05:58 PM
juicysharpie
Local Max/Min Values using Partial Derivatives
f(x, y) = 2x^3 + xy^2 + 5x^2 + y^2 + 8

Find the local max, min and saddle points of the function.

So far, I have:
-partial derivative of x= 6x^2 + y^2 + 10x
-partial derivative of y= 2xy+2y

But I'm really confused what to do next. My book says to set both of the partial derivative equations to 0 and then solve for one of the variables, but I get some weird answer that doesn't work in both equations.

Thanks so much for your help! (Nod)
• May 18th 2009, 06:03 PM
TheEmptySet
Quote:

Originally Posted by juicysharpie
f(x, y) = 2x^3 + xy^2 + 5x^2 + y^2 + 8

Find the local max, min and saddle points of the function.

So far, I have:
-partial derivative of x= 6x^2 + y^2 + 10x
-partial derivative of y= 2xy+2y

But I'm really confused what to do next. My book says to set both of the partial derivative equations to 0 and then solve for one of the variables, but I get some weird answer that doesn't work in both equations.

Thanks so much for your help! (Nod)

So you need to solve this system of equaitons

$\displaystyle 6x^2+y^2+10x=0$ and

$\displaystyle 2xy+2y=0$

Notice that we can factor the bottom equaiton to get

$\displaystyle 2y(x+1)=0$ so by the zero factor principle we get

either $\displaystyle 2y=0$ or $\displaystyle x+1=0$

So now if $\displaystyle y=0$ we can plug that into the top equation to get $\displaystyle 6x^2+10x=0 \iff 2x(3x+5)=0$ si we get two possible ordered pairs $\displaystyle (0,0) \mbox{ or } (-\frac{5}{3},0)$

You need to use the other solution for x above to find the rest of your critical points. I hope this helps
• May 18th 2009, 07:39 PM
arbolis
The first step is as the book says and as TheEmptySet started you on. Once you have all the critical points you're not done! You have to show whether they are local max/min or saddle point.
Call $\displaystyle D=f_{xx}(a,b)f_{yy}(a,b)-f_{xxy}(a,b)$.
Now, if $\displaystyle D>0$ and $\displaystyle f_{xx}(a,b)>0$, then $\displaystyle (a,b)$ is a local minimum.
If $\displaystyle D>0$ and $\displaystyle f_{xx}(a,b)<0$ then $\displaystyle (a,b)$ is a local maximum.
If $\displaystyle D<0$ then $\displaystyle (a,b)$ is a saddle point.

Finally, if $\displaystyle D>0$ and $\displaystyle f_{xx}(a,b)=0$ then you cannot conclude using this formula, and I don't really know how to proceed. I hope your exercise doesn't fall into this category.