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# help

0
261
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+1200

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
+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
+1200
+1

Thanks Rom

Jul 29, 2019