April 21st, 2016, 02:27 AM  #1 
Newbie Joined: Apr 2015 From: Slovakia Posts: 5 Thanks: 0  Local extrema
Hi, I have to find the local extrema of the function 20160421 12_11_32IMA2_9.png . But the problem is that as far as I know I have to find critical numbers first, but after I find first derivative. I get f'(x) = e^(3x+2y) *(6x6y+8y^2); what should I do now? If I set it equal to 0, I am kind of lost. Last edited by Mrto; April 21st, 2016 at 02:55 AM. 
April 21st, 2016, 03:36 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,762 Thanks: 697 
Your description is confusing. are x and y separate variables? In that case you have a surface, so it is not clear what you are looking for.

April 23rd, 2016, 10:08 AM  #3 
Newbie Joined: Apr 2015 From: Slovakia Posts: 5 Thanks: 0 
Idk, its homework and the assignment only says "Find local extrems of function (function in op)" . We did simillar problem to this, we first derivated with respect to x, then with respect to y, then set both equal to 0 and found critial numbers. And then I dont remember and i dont have my book near me right now .

April 23rd, 2016, 03:26 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,762 Thanks: 697 
You want $\displaystyle \frac{\partial f(x,y)}{\partial x}\ and\ \frac{\partial f(x,y)}{\partial y}$ To save writing let $\displaystyle g(x,y)=e^{3x+2y}\ and\ p(x,y)=3x^26xy+6y^2$ You need to compute $\displaystyle \frac{\partial f(x,y)}{\partial x}=p(x,y)\frac{\partial g(x,y)}{\partial x}+g(x,y)\frac{\partial p(x,y)}{\partial x}\ and\ \frac{\partial f(x,y)}{\partial y}=p(x,y)\frac{\partial g(x,y)}{\partial y}+g(x,y)\frac{\partial p(x,y)}{\partial y}$. Since $\displaystyle \frac{\partial g(x,y)}{\partial x}=3g(x,y)\ and\ \frac{\partial g(x,y)}{\partial y}=2g(x,y)$, when you set things = 0, g(x,y) is a common factor and can be dropped. You should be able to do this. 

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extrema, local 
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