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Part CCSIR NET December 2025the-isoperimetric-constraint-alone-pins-down-which-power-of-x-is-the-extremal

The isoperimetric constraint alone pins down which power of x is the extremal

Suppose y(x) is the extremal of the variational problem J(y)=∫01x2(y′)2J(y) = \int_{0}^{1} x^{2}(y')^{2} dx subject to y(0)=0,y(1)=1,∫01y2y(0)=0, y(1)=1, \int_{0}^{1}y^{2}dx = 1/7. Then which of the following statements are true?

  1. A.y′(1/2) = 3/4✓
  2. B.y′(1/3)=1y'(1/\sqrt{3}) = 1✓
  3. C.y′(1/3) = 1/3✓
  4. D.y′(1/4) = 1/2

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward.

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More on this topic

The chapter behind this: Constrained variational problems — free to read

From Calculus of Variations › Isoperimetric problems

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