If $$m,n$$ are negative real numbers such that $$y = \dfrac{x}{(x-m)(x-n)} $$, find the maximum value of $$y$$ in the domain of $$x0 $$

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If $$m,n$$ are negative real numbers such that $$y = \dfrac{x}{(x-m)(x-n)} $$, find the maximum value of $$y$$ in the domain of $$x>0 $$

Uniteasy Admin · Accepted Answer

Taking reciprocal of the given equation gives
$$\dfrac{1}{y}=\dfrac{(x-m)(x-n)}{x}=\dfrac{x^2-(m+n)x+mn)}{x}   $$
$$=-(m+n)+x+\dfrac{mn}{x} $$
$$=-(m+n)+\Big( \sqrt{x}-\dfrac{\sqrt{mn} }{\sqrt{x} }  \Big) ^2+2\sqrt{x}\cdotp \dfrac{\sqrt{mn} }{\sqrt{x} } $$
$$\geq -(m+n)+\sqrt{mn} $$
Only when $$x=\dfrac{mn}{x} $$, that is $$x=\sqrt{mn} $$, $$ \dfrac{1}{y}$$ has minimum value of $$-(m+n)+\sqrt{mn}$$
Taking reciprocal of $$\dfrac{1}{y}$$ gives maximum value of $$y$$
$$y_{max} = \dfrac{1}{-(m+n)+\sqrt{mn}} =\dfrac{1}{(\sqrt{-m}+\sqrt{-n}  )^2}  $$
To verify the result, let $$m=-0.1, n=-0.2$$
The original function is 
$$y = \dfrac{x}{(x+0.1)(x+0.2)} $$
Plot the curve of the function. It's found that $$y$$ has maximum value when $$x>0$$. Drag the dynamic point. It's found the maximum value shown on the curve is consistent to the value calculated using the derived result 1.715.

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