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