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Given that \(xy = \dfrac32\) and both x and y are nonnegative real numbers, find the minimum value of \(10x + \dfrac{3y}5.\)

Guest Jan 27, 2018

Best Answer 

 #1
avatar+92179 
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Given that  \(xy = \dfrac32 \)   and both x and y are nonnegative real numbers, find the minimum value of 

\(10x + \dfrac{3y}5\)

 

 

\(y=\frac{3}{2x}\)

 

so

\(​​​​z=10x+\frac{3y}{5}\\ ​​​​z=10x+\frac{3}{5}\times \frac{3}{2x}\\ ​​​​z=10x+\frac{9}{10x}\\ ​​​​z=10x+0.9x^{-1}\\ \frac{dz}{dx}=10-0.9x^{-2}\\ \frac{d^2z}{dx^2}=1.8x^{-3}>0\qquad \text{So any stationary point will be a minimum.}\\ \text{Find stat point}\\ 10-0.9x^{-2}=0\\ 10=0.9x^{-2}\\ 10=\frac{9}{10x^2}\\ 100x^2=9\\ x^2=\frac{9}{100}\\ x=\frac{3}{10}\qquad \text{Since x>0}\\ so\\ ​​​​z=10x+0.9x^{-1}\\ min\;\;value\\ =10*\frac{3}{10}+\frac{9*10}{10*3}\\ =3+3\\ =6 \)

 

 

check:

 

Here is the graph:

Melody  Jan 28, 2018
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1+0 Answers

 #1
avatar+92179 
+2
Best Answer

Given that  \(xy = \dfrac32 \)   and both x and y are nonnegative real numbers, find the minimum value of 

\(10x + \dfrac{3y}5\)

 

 

\(y=\frac{3}{2x}\)

 

so

\(​​​​z=10x+\frac{3y}{5}\\ ​​​​z=10x+\frac{3}{5}\times \frac{3}{2x}\\ ​​​​z=10x+\frac{9}{10x}\\ ​​​​z=10x+0.9x^{-1}\\ \frac{dz}{dx}=10-0.9x^{-2}\\ \frac{d^2z}{dx^2}=1.8x^{-3}>0\qquad \text{So any stationary point will be a minimum.}\\ \text{Find stat point}\\ 10-0.9x^{-2}=0\\ 10=0.9x^{-2}\\ 10=\frac{9}{10x^2}\\ 100x^2=9\\ x^2=\frac{9}{100}\\ x=\frac{3}{10}\qquad \text{Since x>0}\\ so\\ ​​​​z=10x+0.9x^{-1}\\ min\;\;value\\ =10*\frac{3}{10}+\frac{9*10}{10*3}\\ =3+3\\ =6 \)

 

 

check:

 

Here is the graph:

Melody  Jan 28, 2018

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