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Application of Derivatives question

2006 · Shift 0 · Q58
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Application of Derivatives question

2006 · Shift 0 · Q58

JEE MainMathematicsApplication of DerivativesMCQ+4 / −1
A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fence are of same length xxx. The maximum area enclosed by the park is
  1. A
    32x2{3 \over 2}{x^2}23​x2
  2. B
    x38\sqrt {{{{x^3}} \over 8}}8x3​​
  3. C
    12x2{1 \over 2}{x^2}21​x2
  4. D
    πx2\pi {x^2}πx2
View written solutionFree

Correct answer: C

  1. Interpret the geometry

A triangular park is formed with:

  • two equal fenced sides, each of length xxx,
  • the third side along the river bank.

So the triangle is an isosceles triangle with two equal sides xxx and included angle θ\thetaθ between them.

  1. Write the area in terms of θ\thetaθ

For a triangle with two sides a,ba,ba,b and included angle θ\thetaθ,

A=12absin⁡θA=\frac12 ab\sin\thetaA=21​absinθ

Here a=b=xa=b=xa=b=x, so

A=12x⋅xsin⁡θ=12x2sin⁡θA=\frac12 x\cdot x\sin\theta=\frac12 x^2\sin\thetaA=21​x⋅xsinθ=21​x2sinθ

  1. Maximize the area

Since xxx is fixed, maximizing AAA means maximizing sin⁡θ\sin\thetasinθ.

We know

sin⁡θ≤1\sin\theta \le 1sinθ≤1

with equality when

θ=90∘\theta=90^\circθ=90∘

Hence the maximum area is

Amax⁡=12x2⋅1=12x2A_{\max}=\frac12 x^2\cdot 1=\frac12 x^2Amax​=21​x2⋅1=21​x2

  1. Check the options
  • A: 32x2\dfrac32 x^223​x2 — too large, not possible.
  • B: x38\sqrt{\dfrac{x^3}{8}}8x3​​ — dimensionally incorrect for area.
  • C: 12x2\dfrac12 x^221​x2 — correct.
  • D: πx2\pi x^2πx2 — not related to triangle area here.

Therefore, the correct option is C.

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