+0  
 
+1
272
1
avatar

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

 Jan 27, 2018

Best Answer 

 #1
avatar+96949 
+2

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:

 Jan 28, 2018
 #1
avatar+96949 
+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

26 Online Users

avatar
avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.