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# help

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Find the maximum value of $$\frac{x - y}{x^4 + y^4 + 6}$$ over all real numbers x and y.

Mar 3, 2019

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$$f=\dfrac{x-y}{x^4+y^4+6}$$

$$\text{intuition tells me }y=-x\\ \text{plugging this in we get}\\ f = \dfrac{2x}{2x^4+6} = \dfrac{1}{x^3+\frac 3 x}\\ \dfrac{df}{dx} = \dfrac{1}{\left(x^3+\frac 3 x\right)^2}\left(3x^2-\dfrac{3}{x^2}\right)\\ \dfrac{df}{dx}=0 \Rightarrow 3x^2=\dfrac{3}{x^2}\\ x^4=1,~x=\pm 1,~y=-x\\ x=-1,~y=1 \text{ will be a minimum}\\ x=1,~y=-1 \text{ will be a maximum}$$

$$\text{The maximum value is }\dfrac{2}{2+6}=\dfrac 1 4$$

This can all be verified by actually solving $$\nabla f = (0,0)$$ for it's real roots.  These turn out to be as mentioned above.

Mar 3, 2019