My Math Forum Critical point

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July 13th, 2017, 08:58 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 09:02 AM.

 July 13th, 2017, 09:11 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,656 Thanks: 1327 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, 09:14 AM #3 Senior Member   Joined: Oct 2009 Posts: 142 Thanks: 60 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, 12:00 PM   #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.

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