My Math Forum Local extrema

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 April 21st, 2016, 03:27 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 2016-04-21 12_11_32-IMA2_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) *(6x-6y+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 03:55 AM.
 April 21st, 2016, 04:36 PM #2 Global Moderator   Joined: May 2007 Posts: 6,684 Thanks: 659 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, 11:08 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, 04:26 PM #4 Global Moderator   Joined: May 2007 Posts: 6,684 Thanks: 659 You want $\displaystyle \frac{\partial f(x,y)}{\partial x}\ and\ \frac{\partial f(x,y)}{\partial y}$ To save writing let $\displaystyle g(x,y)=e^{3x+2y}\ and\ p(x,y)=3x^2-6xy+6y^2$ 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. Thanks from Mrto

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