Minimum of a function $f(x,y)=\frac{(1+2y)(1+\frac{x}{2})}{(1+y)(1+x)+x}$

Question

what is the minimum of a function \begin{align} f(x,y)&=\frac{(1+2y)(1+\frac{x}{2})}{(1+y)(1+x)+x}\\ \text {s.t. }& 1 \le y \le x \le y(1+y) \end{align}

I asked Wolfram and Alfa and it says that the minimum is $f(x,y)=0.8565$. I found the derivative with respect to both x and y: \begin{align} \frac{df(x,y)}{dx}&=\frac{4\, x + 3\, y + 4\, x\, y + 2\, x^2\, y + 4\, x^2 + 1}{2\, {\left(2\, x + y + x\, y + 1\right)}^2}\\ \frac{df(x,y)}{dy}&=- \frac{\left(2\, x + 1\right)\, \left(\frac{x}{2} + 1\right)\, \left(x + 1\right)}{{\left(x + \left(x + 1\right)\, \left(y + 1\right)\right)}^2} \end{align}

Should I set up Lagrangian next or is there a better way to solve this?

Answer 1

hint:

$f(x,y)=\dfrac{(\frac{1}{2}+y)x+(y+1)+1}{(2+y)x+(y+1)}=1+\dfrac{1-\frac{3}{2}x}{(2+y)x+(y+1)}$

so it is a mono decreasing function for $x \implies f_{min}=f(y(y+1),y)=f_1(y)$

now use same method to check if $f_1(y)$ is mono decreasing function or not.thenthe answer is easy to get.

Answer 2

i think first you should obtain grad of f then grad f=0 if you find x,y that Applies to domain its minimum else using lagrangian method.