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

Guest Mar 3, 2019

#1**+1 **

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

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

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

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.

Rom Mar 3, 2019