Question:
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 :
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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Write y in terms of x using the line, then maximise \(xy^4\) (calculus or weighted AM-GM).
Updated On: Oct 31, 2026
- y = 4x
- x = 4y
- y = 4x + \( \alpha \)
- x = 4y \( - \) \( \alpha \)
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The Correct Option is A
Solution and Explanation
Step 1: The line through \((50+\alpha,0)\) and \((0,50+\alpha)\) is \(x+y=50+\alpha\). Let \(s=50+\alpha\).
Step 2: \(f(x)=x(s-x)^4\Rightarrow f'(x)=(s-x)^4-4x(s-x)^3=(s-x)^3(s-5x)\).
Step 3: For \(0<x<s\), f' changes sign from + to - at \(x=s/5\), so the maximum is at \(x=\dfrac s5\), \(y=\dfrac{4s}5\).
Step 4: Then \(y=4x\), so the point lies on the line \(y=4x\) (the other lines hold only for special alpha).
Answer: (A) y = 4x
Step 2: \(f(x)=x(s-x)^4\Rightarrow f'(x)=(s-x)^4-4x(s-x)^3=(s-x)^3(s-5x)\).
Step 3: For \(0<x<s\), f' changes sign from + to - at \(x=s/5\), so the maximum is at \(x=\dfrac s5\), \(y=\dfrac{4s}5\).
Step 4: Then \(y=4x\), so the point lies on the line \(y=4x\) (the other lines hold only for special alpha).
Answer: (A) y = 4x
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