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Given that xy = 1/2  and both x and y are nonnegative real numbers, find the minimum value of 4x + 9y.

 Oct 13, 2020
 #1
avatar+10324 
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Given that xy = 1/2  and both x and y are nonnegative real numbers, find the minimum value of 4x + 9y.

 

Hello Guest!

 

\(xy=\frac{1}{2} \)       \(y=\frac{1}{2x}\)

 

\(f(x)=4x+9y\\ f(x)=4x+\frac{9}{2x}\\ f(x)=4x+4.5x^{-1}\)

\(\color{blue}\frac{df(x)}{dx}=4-4.5x^{-2}=0\\ \frac{4.5}{x^2}=4\\ x^2=\frac{4.5}{4}\)

\(x=1.06066\)

\(y=0.47140\)

\(4x+9y=8.48528\)

 

The minimum value of 4x + 9y is 8.48528.

laugh  !

 
 Oct 13, 2020
 #2
avatar+10324 
0

Given that xy = 1/2  and both x and y are nonnegative real numbers, find the minimum value of 4x + 9y.

 

 \(xy=\frac{1}{2}\)       \(x=\frac{1}{2y}\)

 

\(f(y)=4x+9y\\ f(y)=\frac{2}{y}+9y=2y^{-1}+9y\)

\(\color{blue}\frac{df(y)}{dy}=-2y^{-2}+9=0\\ \frac{2}{y^2}=9\\ y^2=\frac{2}{9}\)

\(y=0.42140\\ x=1.06066\)

 

       ⇓

\(4x+9y=8.48528 \)

 

q.e.d.

laugh  !

 
 Oct 13, 2020

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