Given two positive real numbers x,y so that x+y≥10x+y≥10 Find the smallest value of P , know P = 2x+y+30/x+5/y

Question by Phan Huy Toàn

16/08/2017 at 15:44

Find the smallest value of P , know P = $2 x + y + \frac{30}{x} + \frac{5}{y}$

Kaya Renger Coordinator 16/08/2017 at 15:49

Apply Cauchy inequality , we have :

$P=2x+y+\dfrac{30}{x}+\dfrac{5}{y}=\dfrac{4}{5}\left(x+y\right)+\left(\dfrac{6x}{5}+\dfrac{30}{x}\right)+\left(\dfrac{y}{5}+\dfrac{5}{y}\right)$

$P\ge\dfrac{4}{5}.10+2.\sqrt{\dfrac{6x}{5}.\dfrac{30}{x}}+2.\sqrt{\dfrac{y}{5}.\dfrac{5}{y}}=8+6+2=16$

So $Min_P=16$

<=> x = y = 5
Selected by MathYouLike
Kaya Renger Coordinator 16/08/2017 at 15:52

That is my question ok : https://e-learning.codienhanoi.edu.vn/questions/1902.html

Searching before asking , ok cheater :V
Help you solve math 16/08/2017 at 15:54

P=4/5(x+y)+(6/5x+30/x)+(y/5+5/y)≥45.10+2√6/5x.30/x+2√y/5.5/y

=8+12+2=22=8+12+2=22

minP=22⇔x=y=5⇔x=y=5