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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Official solution
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=xy4
\[ = (50 + \alpha - y){y^4}\]
\[ = (50 + \alpha ){y^4} - {y^5}\]
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