Finding minimum in functions of 2 variables

Question

Question : Find all (x,y), x,y> 0 for which f(x,y) is minimum where $f(x,y) = \frac{x^4}{y^4}+\frac{y^4}{x^4}-\frac{x^2}{y^2}-\frac{y^2}{x^2}+\frac{x}{y}+\frac{y}{x}$

My attempt:The values of (x,y) for which $\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=0$ are critical points

$\frac{\partial f}{\partial x}=\frac{4x^3}{y^4}-\frac{4y^4}{x^5}-\frac{2x}{y^2}-\frac{2y^2}{x^3}+\frac{1}{y}-\frac{y}{x^2}=0$

$\frac{\partial f}{\partial y}=-\frac{4x^4}{y^5}+\frac{4y^3}{x^4}-\frac{2x^2}{y^3}-\frac{2y}{x^2}-\frac{x}{y^2}+\frac{1}{x}=0$

Now I got stuck on solving these equations. It seems complicated. Kindly help me.

Answer 1

Here it goes,

Let $$\frac{x}{y}=t$$

$$f(t)=t^4+\frac{1}{t^4}+t+\frac{1}{t}-(t^2+\frac{1}{t^2})$$

Now

$$t^4+\frac{1}{t^4}=(t^2+\frac{1}{t^2})^2-2=((t+\frac{1}{t})^2-2)^2-2=((t+\frac{1}{t})^4-4(t+\frac{1}{t})^2+4-2)=((t+\frac{1}{t})^4-4(t+\frac{1}{t})^2+2)$$

$$t^2+\frac{1}{t^2}=(t+\frac{1}{t})^2-2$$

Let $$t+\frac{1}{t}=z$$

$$t^4+\frac{1}{t^4}=z^4-4z^2+2$$ $$t^2+\frac{1}{t^2}=z^2-2$$

$$f(t)=f(z)=z^4-4z^2+2+z-(z^2-2)=z^4-5z^2+z+4$$

Now $$z\geq2$$ as $$t+\frac{1}{t}\geq2$$

$$f'(z)=4z^3-10z+1$$ $$f''(z)=12z^2-10$$ For $z\geq2$,$f''(z)>0$,Which means, f'(z) is increasing function for $z\geq2$

Now $$f'(2)=13>0$$ Which means $f'(x)>0$ for $x\geq2$ because it is increasing, Which means $f(x)$ is increasing for $z\geq2$

hence

$$f(z)_{min}=f(2)=2=f(x,y)_{min}$$

Answer 2

Observe that $f(x,y)$ and its derivatives with respect to $x$ and $y$ are symmetric in $x,y$, which suggests that you satisfy $\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = 0$ with $y = x$ (for $x,y > 0$). The second derivative of $f(x,y)$ with respect to $x$ and $y$ is positive for $y = x$ (and $x,y > 0$), so you have indeed infinitely many minima along $y = x$.

Answer 3

Let us substitute $\displaystyle t=\frac xy+\frac yx\ge 2\sqrt{\frac xy\cdot\frac yx}=2$, and the equality $t=2$ is obtained if and only if $x=y$. Then the given function is a function of $t$, $$ \begin{aligned} t^4 &= &\frac {x^4}{y^4} \qquad+ 4\frac {x^2}{y^2} \qquad+&\ 6 \qquad+4\frac {y^2}{x^2}\qquad + \frac {y^4}{x^4}\ ,\\ t^2 &= &\frac {x^2}{y^2} \qquad+&\ 2\qquad +\frac {y^2}{x^2}\ ,\\ t&= &\frac xy \ +&\ \frac yx\ ,\\ h(t)=t^4-5t^2+t+4&=&f(x,y)=&\ f(y,x)\ , \end{aligned} $$ and the help function $h$ of $t$ above is strictly increasing on $[2,\infty)$ because its derivative $h'(t)=4t^3-10t+1$ has three real roots in the intervals $(-\infty,-1)$, $(-1,1)$, and $(1,2)$, (for this we compute $f(-1)=7>0$, $f(1)=-5<0$, $f(2)=13>0$) so $h'$ has the plus sign on the interval $[2,\infty)$, so $h$ is strictly increasing on $[,\infty)$.

Tracing back this information to $f$, we see that each point $x,y$ with $\frac xy+\frac yx=2$ is a point with an absolutely minimal value for $f$, this value being $f(x,x)=1+1-1-1+1+1=2$, or alternatively $h(2)= 2^4-5\cdot2^2+2+4=2$.

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Here is a contour plot of the function $f$ on a small square: