Find the absolute maximum and minimum values of the function $$f(x, y) = x^{2} + 3y^{2} - y$$ over the region $x^{2} + 2y^{2} \leq 1$
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Lagrange multipliers – Jürgen Sukumaran Feb 16 '18 at 12:44
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1Welcome to stackexchange. Your question got lots of downvotes and a vote to close because you showed no effort of your own. Please do that if you want to ask further questions here. You will learn more that way.In the future, do not post images. Use mathjax for format mathematics: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Ethan Bolker Feb 16 '18 at 12:59
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it fist you must look of the extrema in the inner of the region: $$f(x,y)=x^2+3y^2-y$$ at first we have $$\frac{\partial f(x,y)}{\partial x}=2x$$ $$\frac{\partial f(x,y)}{\partial y}=6y-1$$ and then the extrema on the boundary $$x^2+2y^2=1$$ for the second step consider the function $$g(y)=1-2y^2+3y^2-y$$ you will get the maximimum: $$\frac{3}{2}+\frac{1}{\sqrt{2}}$$ for $$x=0,y=-\frac{1}{\sqrt{2}}$$ and the minimum: $$-\frac{1}{12}$$ for $$x=0,y=\frac{1}{6}$$
Dr. Sonnhard Graubner
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2I'm disappointed that you chose to answer this question from a new user only five minutes after it was asked, rather than encouraging the OP to provide his or her thoughts first. – Ethan Bolker Feb 16 '18 at 12:55
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ok i will remember this in the future! thanks for your hint! – Dr. Sonnhard Graubner Feb 16 '18 at 12:57