# What is the maximum value of $\frac{7x+2y}{2x+2}+\frac{3x+8y}{2y+2}$ while $0≤x,y≤1$

Find maximum value of $$\left(\dfrac{7x+2y}{2x+2}+\dfrac{3x+8y}{2y+2}\right)$$ for $$0≤x,y≤1$$

My approach was to use A.M-G.M inequality or cauchy shbert inequality, but I failed.

• The same “trick” as in this answer math.stackexchange.com/a/3476058/42969 to your previous question works here: Decrease both denominators to $2x + 2y$. Dec 15, 2019 at 7:50
• oh. I see. Then the answer would be 5. Dec 15, 2019 at 7:52
• Should I delete the question, or answer it myself? Dec 15, 2019 at 7:52

Since $$x,y\leq1$$ $$\dfrac{7x+2y}{2x+2}\leq\dfrac{7x+2y}{2x+2y}\\ \dfrac{3x+8y}{2y+2}\leq\dfrac{3x+8y}{2x+2y}$$ so $$\dfrac{7x+2y}{2x+2}+\dfrac{3x+8y}{2y+2}\leq\dfrac{10x+10y}{2x+2y}=5$$
Equality will hold when $$x=1$$ & $$y=1$$
• One small point is you only showed that $5$ is an upper bound. For a complete answer, since the question asks for the maximum value, you need to show this value can actually reached. Note this is done for the case in your linked answer. For the case here, the value of $5$ occurs, for example, when $x = y = 1$, so you should explicitly mention this. Dec 15, 2019 at 8:07
$$f(x,y) = \frac{7x+2y}{2x+2} + \frac{3x+8y}{2y+2} = \left( \frac 72 - \frac{7-2y}{2x+2} \right)+ \left( 4 - \frac{8 - 3x}{2y+2} \right)$$ is strictly increasing in both variables on $$[0, 1]^2$$, therefore $$f(x, y) \le f(1,1) = 5$$ with equality exactly for $$(x, y) = (1,1)$$.