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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.

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

2 Answers 2

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I found a solution myself with helps of the comments above on the question.
The solution is similar with this solution of my previous question.
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 $x=1$ & $y=1$

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    $\begingroup$ 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. $\endgroup$ Dec 15, 2019 at 8:07
  • $\begingroup$ @JohnOmielan I see. I will edit the answer in that way. $\endgroup$
    – dissolve
    Dec 15, 2019 at 9:10
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$$ f(x,y) = \frac{7x+2y}{2x+2} + \frac{3x+8y}{2y+2} = \left( \frac 72 - \frac{7-2y}{2x+2} \right)+ \left( 4 - \frac{8 - 3x}{2y+2} \right) $$ is strictly increasing in both variables on $[0, 1]^2$, therefore $$ f(x, y) \le f(1,1) = 5 $$ with equality exactly for $(x, y) = (1,1)$.

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