﻿ If m,n are negative real numbers such that y = \dfrac{x}{(x-m)(x-n)} , find the maximum

#### Question

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

Collected in the board: Inequality

Steven Zheng posted 1 year ago

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.

Steven Zheng posted 1 year ago

Scroll to Top