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

Solution

Maths sir · 4 steps

Use the line to write x in terms of y, then differentiate a function of one variable.

1. Equation of line. Both intercepts equal k = 50 + α, so the line is x + y = k. Between the intercepts x, y > 0, and there x = k - y.
2. One variable. Let f(y) = (k - y)y⁴. Differentiate and set the derivative to zero, keeping 0 < y < k.

   f'(y) = 4y³(k - y) - y⁴ = y³(4k - 5y) = 0 ⇒ y = 4k/5

3. Check maximum. f''(y) = 12ky² - 20y³ = 4y²(3k - 5y). At y = 4k/5 this is -4ky² < 0, so it is a maximum. Then x = k - 4k/5 = k/5.
4. So. y = 4k/5 = 4 × k/5 = 4x. The point lies on y = 4x, which is option (a).

The idea to keep. To maximise xᵐ yⁿ when x + y is fixed, split the sum in the ratio of the powers: x : y = m : n. AM-GM on x + y/4 + y/4 + y/4 + y/4 gives the same result.

Answer from the NTA final answer key. Solution by PrepWiser.