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  • 4 years ago

I am working on critical points and the second derivative test. I understand the calculus but am having difficulty with the algebra. How do you solve the partial derivative fx=y^2+2xy+8y and fy=2xy+x^2+8x for 0 to find the critical points. I already have one solution of (0,0) are there any more critical points and techniques to solve this?

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  1. anonymous
    • 4 years ago
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    \[f _{x}=y ^{2}+2xy+8y=0\]factor\[y(y+2x+8)=0\]You need to put both of these situations into the other partial derivative's solutions. \[y=0\]\[y+2x+8=0\]Now work with the partial with respect to y for a moment. \[f _{y}=2xy+x ^{2}+8x=0\]factor\[x(2y+x+8)=0\]Will give\[x=0\]\[2y+x+8=0\]Both of these situations will need to be combined with the first set. Essentially, every solution needs to be combined with every other solution. From these, it is easy to see what you already got... (0,0)... now check the other combinations. x=0 into y+2x+8=0 y+2(0)+8=0 y=-8 giving the point (0,-8) Now put y=0 into 2y+x+8=0 2(0)+x+8=0 x=-8 giving the point (-8,0) Now solve the system y+2x+8=0 and 2y+x+8=0 this will result in (-8/3, -8/3) I worked backwards to get something resembling your original function and graphed it in some nice 3d graphing software that I use. (0,0), (0,-8), and (-8,0) are saddles (-8/3, -8/3) is a max. for \[f(x,y)=xy ^{2}+x ^{2}y+8xy+c\]You should be able to verify this using the second derivative test. Get you java plug-in up and running then check out http://web.monroecc.edu/manila/webfiles/calcNSF/JavaCode/CalcPlot3D.htm

  2. anonymous
    • 4 years ago
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    Thank you very much!! That helped a lot. It turns out it was not as difficult as I thought. Thanks again!

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