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July 13th, 2017, 07:58 AM   #1
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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
Attached Images Screen Shot 2017-07-13 at 11.49.34 AM.jpg (14.7 KB, 7 views)

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: 3,002 Thanks: 1587 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) Thanks from SenatorArmstrong July 13th, 2017, 08:14 AM #3 Senior Member   Joined: Oct 2009 Posts: 850 Thanks: 326 You made some non-essential 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. Thanks from SenatorArmstrong July 13th, 2017, 11:00 AM   #4
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Quote:
 Originally Posted by Micrm@ss You made some non-essential 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.
Thank you for responding. $y=0$ ... Slipped right over my head. But I will be on the look out in the future for situations where I have to consider multiple cases. Tags critical, point Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post puppypower123 Calculus 4 March 13th, 2017 12:22 PM MFP Calculus 6 August 21st, 2013 02:42 PM felicia184 Calculus 5 September 17th, 2012 10:05 AM Mike86 Applied Math 2 October 9th, 2010 06:02 AM felicia184 Algebra 0 December 31st, 1969 04:00 PM

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