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Find the minimum value of $$\frac{(x + y)^3}{x^2 y}$$, where x and y are positive real numbers.

Nov 12, 2019

#1
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Find the minimum value of $$\dfrac{(x + y)^3}{x^2 y}$$, where x and y are positive real numbers.

$$\begin{array}{|rcll|} \hline \mathbf{f(x,y)} & \mathbf{=} & \mathbf{\dfrac{(x + y)^3}{x^2 y} } \\\\ \dfrac{\partial f(x,y)} {\partial x} &=& \dfrac{(x + y)^3}{x^2 y}\left( \dfrac{3(x+y)^3}{(x+y)^3} - \dfrac{2xy}{x^2y} \right) \\ &=& \dfrac{(x + y)^3}{x^2 y}\left( \dfrac{3}{x+y} - \dfrac{2}{x} \right) \\ 0 &=& \dfrac{(x + y)^3}{x^2 y}\left( \dfrac{3}{x+y} - \dfrac{2}{x} \right) \\ \dfrac{3}{x+y} - \dfrac{2}{x} &=& 0 \\ \dfrac{3}{x+y} &=& \dfrac{2}{x} \\ \dfrac{x+y}{3} &=& \dfrac{x}{2} \\ 2x+2y &=& 3x \\ \mathbf{x} &=& \mathbf{2y} \\\\ \dfrac{\partial f(x,y)} {\partial y} &=& \dfrac{(x + y)^3}{x^2 y}\left( \dfrac{3(x+y)^3}{(x+y)^3} - \dfrac{x^2}{x^2y} \right) \\ &=& \dfrac{(x + y)^3}{x^2 y}\left( \dfrac{3}{x+y} - \dfrac{1}{y} \right) \\ 0 &=& \dfrac{(x + y)^3}{x^2 y}\left( \dfrac{3}{x+y} - \dfrac{1}{y} \right) \\ \dfrac{3}{x+y} - \dfrac{1}{y} &=& 0 \\ \dfrac{3}{x+y} &=& \dfrac{1}{y} \\ \dfrac{x+y}{3} &=& \dfrac{y} {1} \\ x+y &=& 3y \\ \mathbf{x} &=& \mathbf{2y} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{f(x,y)} & \mathbf{=} & \mathbf{\dfrac{(x + y)^3}{x^2 y} } \quad | \quad \text{minimize } x=2y \\\\ f(x,y) &=& \dfrac{(2y + y)^3}{(2y)^2 y} \\\\ &=& \dfrac{(3y)^3}{4y^2y} \\\\ &=& \dfrac{27y^3}{4y^3} \quad | \quad y \text{ is a positive real number} \\\\ &=& \dfrac{27}{4} \\\\ \mathbf{f(x,y)_{\text{minimum}}} &=& \mathbf{6.75} \\ \hline \end{array}$$

Nov 13, 2019