Max. and Min. value of $f(x,y) = \frac{x-y}{x^4+y^4+6}$

Question

[1] Calculation of Max. and Min. value of $\displaystyle f(x) = \sqrt{x^3-6x^2+21x+18}$, where $\displaystyle -\frac{1}{2}\leq x\leq 1$

[2] Calculation of Max. and Min. value of $\displaystyle f(x,y) = \frac{x-y}{x^4+y^4+6}\;,$ where $x,y\in \mathbb{R}$

$\bf{My\; Try::(1)}$Let $g(x) = x^3-6x^2+21x+18$, Now Using Derivative Test.

$g^{'}(x)=3x^2-12x+21=3(x^2-4x+7)=3\left\{(x-2)^2+3\right\}>0\;\forall x\in \mathbb{R}$

So $g(x)$ is Strictly Increasing function.

So $g(x)$ is Min. when $\displaystyle x = -\frac{1}{2}$ and $g(x)$ is Max. when $x=1$

$\bf{My\; Try::(2)}$ Let $\displaystyle f(x,y) = z=\frac{x-y}{x^4+y^4+6}.$

Using $\bf{A.M\geq G.M},\;\; $

$x^4+1\geq 2x^2$ and $y^4+1\geq 2y^2\;,$ we get $x^4+y^4+2\geq 2(x^2+y^2)\Rightarrow x^4+y^4+6\geq (x^2+y^2)+4$

Now I did not understand How can I solve it.

Help Required

Thanks.

Answer 1

If you want Max, $x-y>0$, you want Min,$x-y<0$ for Max: $x-y \le |x-y| \le|x|+|y|$

your process can go on: $x^2+1\ge 2|x|,y^2+1\ge2|y| \implies x^4+y^4+6\ge 4(|x|+|y|)\implies \dfrac{x-y}{x^4+y^4+6} \le \dfrac{1}{4}$ when $x=-y=1$

you can get Min in same way.

Answer 2

Hint: Have a look at this for the second problem Second partial derivative test