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Given that xy=3/2 and x both and y are nonnegative real numbers, find the minimum value of 10x+(3y/5).

Jan 27, 2019

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$$x y = \dfrac 3 2,~x,y \in \mathbb{R^+}\\ y = \dfrac{3}{2x}\\ \text{minimize }10x + \dfrac{3y}{5} = \\ 10x + \dfrac{9}{10x}$$

$$\text{Can we use calculus? Assuming yes}\\ \dfrac{d}{dx} \left(10x + \dfrac{9}{10x}\right) = \\ 10 -\dfrac{9}{10x^2}\\$$

$$\text{Setting this equal to zero and solving for }x \text{ we get}\\ 10 = \dfrac{9}{10x^2}\\ 100x^2 = 9\\ 10x = 3\\ x = \dfrac{3}{10}$$

$$\text{We need to ensure this is a minimum by checking the second derivative at this point}\\ \dfrac{d^2}{dx^2}\left(10x+\dfrac{9}{10x}\right) = \left . \dfrac{18}{10x^3} \right |_{x=\frac{3}{10}} = \\ \dfrac{18\cdot 1000}{10\cdot 27} = \dfrac{200}{3} > 0,~\text{so this is a minimum}\\ x = \dfrac{3}{10},~y = \dfrac{3}{2x} =5\\ xy = \dfrac 3 2$$

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Jan 27, 2019