We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

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

 Mar 3, 2019



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


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


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

9 Online Users