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

 Jul 29, 2019
 #1
avatar+6196 
+2

\(y = \dfrac{3}{2x}\\ v=10x+\dfrac 3 5 \cdot \dfrac{3}{2x} = 10x+\dfrac{9}{10x}\\ \dfrac{dv}{dx} = 10 -\dfrac{9}{10x^2}\\ \dfrac{dv}{dx}=0 \Rightarrow 10=\dfrac{9}{10x^2}\\ \dfrac{100}{9}=\dfrac{1}{x^2}\\ x^2 = \dfrac{9}{100}\\ x = \dfrac{3}{10}\)

 

\(\text{We need to confirm this critical point is a minimum}\\ \dfrac{d^2v}{dx^2} = \dfrac{9}{5x^3}\\ \text{$\left. \dfrac{9}{5x^3}\right |_{x=\frac{3}{10}} > 0$ so this is in fact a minimum}\)

 

\(x = \dfrac{3}{10} \Rightarrow \\ y = \dfrac{3}{2\cdot \frac{3}{10}} = 5\\ 10\left(\dfrac{3}{10}\right) + \dfrac{3(5)}{5} = 3+3= 6\)

 

\(\text{The minimum value of $10x+\dfrac{3y}{5},~x,y >0,~xy=\dfrac 3 2$ is 6}\)

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 Jul 29, 2019
 #2
avatar+1200 
+1

Thanks Rom

 Jul 29, 2019

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