JEE Main 2022 — Application Of Derivatives Question with Solution

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

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

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

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

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

Let $$p=x{y}^4$$

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

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

For maximum or minimum value of p,

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

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

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

$$\therefore$$ $$y=0$$

or

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

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

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

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

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

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

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

$$\Rightarrow y=4x$$

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