July 13th, 2017, 07:58 AM  #1 
Senior Member Joined: Nov 2015 From: Alabama Posts: 140 Thanks: 17 Math Focus: Geometry, Trigonometry, Calculus  Critical point
Hello. I am stuck on a problem where I need to find the critical point. My work is attached to this thread. The function is z = yx^5 + xy^5 + xy. My plan of attack was to find the partial derivatives. Also the 2nd partial derivatives. I then would use a determinant to find a value D. While investigating this value and the value of the second partial derivatives, I could then tell if it is a local max, min, or a saddle point. It seems obvious that at (0,0) both first partial derivatives would go to zero. When solving for D I would get 1 = D which D < 0 and therefore this is a saddle point. The problem states that there is only one critical point. But if the problem did not state this, how can I be certain that I am finished hunting for critical points? I would prefer if I could solve the first partial derivative for y, and then plug y into the second partial derivative to solve for x. Then the x and y coordinates I get, I could investigate further. However, I come to the conclusion that y^4 = 1, which forces me to go complex and deviate away from the nature of the problem. If the picture is too small, I invite you to view the image here Last edited by SenatorArmstrong; July 13th, 2017 at 08:02 AM. 
July 13th, 2017, 08:11 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,626 Thanks: 1306 
your image is too small to read ... $z = yx^5 + xy^5 + xy$ $\dfrac{\partial z}{\partial x} = 5yx^4 + y^5 + y$ $\dfrac{\partial z}{\partial y} = x^5 + 5xy^4 + x$ $5yx^4 + y^5 + y = 0$ $y(5x^4+y^4+1) = 0 \implies y = 0$ ... note $5x^4+y^4+1 > 0$ for all $x$ and $y$ $x^5 + 5xy^4 + x=0$ $x(x^4+5y^4+1) = 0 \implies x = 0$ ... same reasoning as above. believe there's a saddle point at (0,0) 
July 13th, 2017, 08:14 AM  #3 
Senior Member Joined: Oct 2009 Posts: 141 Thanks: 59 
You made some nonessential errors. First of all, when you divide by $5y$, you should make sure that $y\neq 0$. So you absolutely need to split up into cases. A case where $y\neq 0$ and a case where $y=0$. Now, you come to the conclusion $5x^4 = y^4  1$. It is absolutely vital that you know that you are working in real analysis and not complex analysis. In REAL analysis, it is impossible for a fourth power to be negative. So you can immediately conclude from the above that since $5x^4$ is always positive, and since $y^4  1$ is always strictly negative, that there is no real solution. And you do only care about REAL solutions here, since this is real analysis. 
July 13th, 2017, 11:00 AM  #4  
Senior Member Joined: Nov 2015 From: Alabama Posts: 140 Thanks: 17 Math Focus: Geometry, Trigonometry, Calculus  Quote:
 

Tags 
critical, point 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Finding the Critical point  puppypower123  Calculus  4  March 13th, 2017 12:22 PM 
Question regarding a critical point formula  MFP  Calculus  6  August 21st, 2013 02:42 PM 
Critical point  felicia184  Calculus  5  September 17th, 2012 10:05 AM 
NonLinear First Order ODE: Critical Point Linearization  Mike86  Applied Math  2  October 9th, 2010 06:02 AM 
Critical point  felicia184  Algebra  0  December 31st, 1969 04:00 PM 