"If xy4 attains maximum value at the point (x, y) on the line passing through the points (50 + $\alpha$, 0) and (0, 50 + $\alpha$), $\alpha$ 0, then (x, y) also lies on the line :"

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Accepted Answer

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Equation of line passing through the point (50 + $\alpha$, 0) and (0, 50 + $\alpha$) is

$$y - 0 = {{50 + \alpha - 0} \over {0 - (50 + \alpha )}}\left( {x - (50 + \alpha )} \right)$$

$\Rightarrow y = - 1\left( {x - (50 + \alpha )} \right)$

$\Rightarrow y = (50 + \alpha ) - x$

$\Rightarrow x + y = 50 + \alpha$

Let $p = x{y^4}$

$= (50 + \alpha - y){y^4}$

$= (50 + \alpha ){y^4} - {y^5}$

For maximum or minimum value of p,

${{dp} \over {dy}} = 0$

$\Rightarrow 4{y^3}(50 + \alpha ) - 5{y^4} = 0$

$\Rightarrow {y^3}[200 + 4\alpha - 5y] = 0$

$\therefore$ $y = 0$

or

$200 + 4\alpha - 5y = 0$

$\Rightarrow y = {4 \over 5}(50 + \alpha )$

$\therefore$ $x = 50 + \alpha - y$

$= 50 + \alpha - {4 \over 5}(50 + \alpha )$

$= 50 + \alpha \left( {1 - {4 \over 5}} \right)$

$= {1 \over 5}(50 + \alpha )$

$\Rightarrow 4x = {4 \over 5}(50 + \alpha ) = y$

$\Rightarrow y = 4x$

$\therefore$ (x, y) lies on the line y = 4x

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If xy⁴ attains maximum value at the point (x, y) on the line passing through the points (50 + $\alpha$, 0) and (0, 50 + $\alpha$), $\alpha$ > 0, then (x, y) also lies on the line :

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If xy4 attains maximum value at the point (x, y) on the line passing through the points (50 + $\alpha$, 0) and (0, 50 + $\alpha$), $\alpha$ > 0, then (x, y) also lies on the line :

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If xy⁴ attains maximum value at the point (x, y) on the line passing through the points (50 + $\alpha$, 0) and (0, 50 + $\alpha$), $\alpha$ > 0, then (x, y) also lies on the line :