<s>[INST] If $x+2y=8$, find the minumum of $x+y+\frac{3}{x}+\frac{9}{2y}$ (x, y ∈ R+) Question : If $x+2y=8$, find the minumum of $x+y+\dfrac{3}{x}+\dfrac{9}{2y}$ $(x, y \in \mathbb{R}^+)$
I tried to use $x=8-2y$, and I thought I can plug in. But I think I'm not doing it well. I thought about AM-GM, and how can I expand this  using  $x+2y=8$?
 [/INST] $$x+2y=8  \implies x = 8 - 2y\tag{1}$$
Let $$f(x, y) = x+y+\frac{3}{x}+\frac{9}{2y} \tag{2}$$
Using $(1)$ in $(2)$,
$$f(x, y) = 8 - 2y +y+\frac{3}{8 - 2y}+\frac{9}{2y} = 8 - y +\frac{3}{8 - 2y}+\frac{9}{2y} \tag{3}$$
Differentiate $(3)$ wrt $y$, set it equal to $0$ and solve for $y$
$$\frac{-9}{2y^2} + \frac{3}{2(4-y)^2} - 1 = 0$$
The real root is $y = 3$ and the second derivative $$\left[\frac{9}{y^3} + \frac{3}{(4-y)^3} \right]_{y = 3} = \frac{10}{3} > 0$$
From $(1)$, $x = 2$ and the minimum value is $f(2, 3) = 8$
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