Find the minimum of $\space\frac{1}{x}+\frac{1}{y}+c\cdot xy\space$ subject to $\space x+y-c=0$

Question

Let $f(x,y):\mathbb{D}\rightarrow\mathbb{R}$ be the function:

$$f(x,y)=\frac{1}{x}+\frac{1}{y}+c\cdot xy\space\space|\space\space c\in(0,\sqrt[4]8)\text{ $\space$constant}$$ $$\mathbb{D}=\{(x,y)\space|\space x>0,\space y>0\}$$

Find the point $P$ where $f$ gets its minimum value, subject to the equality: $$x+y-c=0$$

I tried:

1) Lagrange multipliers. Unfortunately, they don't seem to help since I get equations I cannot solve (5th degree).

2) Substituting $y=c-x$ to $f$ in order to solve $\frac{d}{dx}f(x,c-x)=0$. That wasn't helpful either, from the same reasons.

Final Solution: The final solution should be $(\frac{c}{2},\frac{c}{2})$; However I could not figure out how to find it by myself.

Thanks!

Answer 1

$\begin{array}\\ f(x,y) &=\dfrac{1}{x}+\dfrac{1}{y}+c xy\\ &=\dfrac{c}{x(c-x)}+c x(c-x)\\ &=\dfrac{c+cx^2(c-x)^2}{x(c-x)}\\ &=c\dfrac{1+x^2(c-x)^2}{x(c-x)}\\ f_x(x,y) &= \dfrac{c^3 x^2 - 4 c^2 x^3 + 5 c x^4 - c - 2 x^5 + 2 x}{x^2 (c - x)^2} \quad\text{according to Wolfy}\\ &= \dfrac{(c - 2 x) (c x - x^2 - 1) (c x - x^2 + 1)}{x^2 (c - x)^2} \quad\text{also according to Wolfy}\\ \end{array} $

so $f$ is extreme at $x=\frac{c}{2}$ or $x^2-cx =\pm 1$ or $x^2-cx+c^2/4 =c^2/4\pm 1$ or $(x-c/2)^2 =c^2/4\pm 1$.

If $(x-c/2)^2 =c^2/4 -1 $ then, since $0 \le c \le \sqrt[4]{8} $, $0 \le c^2 \le \sqrt{8} =2\sqrt{2} $ so $c^2/4 \le \dfrac{\sqrt{2}}{2} \lt 1$ so this can not be.

If $(x-c/2)^2 =c^2/4 +1 $ then $x =\dfrac{c}{2}\pm\sqrt{\dfrac{c^2}{4} +1} $. Since $x > 0$ we must have $x =\dfrac{c}{2}+\sqrt{\dfrac{c^2}{4} +1} $.

If $x = \dfrac{c}{2}$, $x = y$ so $x(c-x) =\dfrac{c^2}{4} $ so

$\begin{array}\\ f(x, y) &=c\dfrac{1+x^2(c-x)^2}{x(c-x)}\\ &=c\dfrac{1+c^4/16}{c^2/4}\\ &=\dfrac{4(1+c^4/16)}{c}\\ &=\dfrac{4}{c}+\dfrac{c^3}{4}\\ \end{array} $

If $x =\dfrac{c}{2}+\sqrt{\dfrac{c^2}{4} +1} $, $y =c-x =\dfrac{c}{2}-\sqrt{\dfrac{c^2}{4} +1} \lt 0 $ which is not allowed.

Therefore the extreme value is at $x=y=\dfrac{c}{2}$ where the value is $\dfrac{4}{c}+\dfrac{c^3}{4} $.

Answer 2

We have $$f(x,y)=\frac{c}{xy}+cxy,$$ which is decreasing function of $xy$ on $(0,1)$.

Thus, we need to found a maximal value of $xy$, for which the equation $$t^2-ct+xy=0$$ has positive roots for $0<c<\sqrt[4]{8}.$

Thus, $c^2-4xy\geq0,$ which says $$xy\leq\frac{1}{4}c^2<\frac{1}{\sqrt2}<1.$$

We see that the maximal value of $xy$ it's $\frac{c^2}{4}$ and occurs for $x=y=\frac{c}{2},$ which says $$\min f(x,y)=\frac{c}{\frac{c^2}{4}}+\frac{c^3}{4}=\frac{4}{c}+\frac{c^3}{4}.$$