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Mathematics · Application of Derivatives

JEE Main 2024 — 5 April, Shift 1 — Question 2

Let a rectangle ABCD of sides 2 and 4 be inscribed in another rectangle PQRSP Q R S such that the vertices of the rectangle ABCDA B C D lie on the sides of the rectangle PQRS. Let a and b be the sides of the rectangle PQRS when its area is maximum. Then (a+b)2(\mathrm{a}+\mathrm{b})^{2} is equal to :

  1. Option A:

    72

    Correct
  2. Option B:

    60

  3. Option C:

    80

  4. Option D:

    64

Answer: A

Step-by-step solution

Area =(4cos⁡θ+2sin⁡θ)(2cos⁡θ+4sin⁡θ)=(4 \cos \theta+2 \sin \theta)(2 \cos \theta+4 \sin \theta)

=8cos⁡2θ+16sin⁡θcos⁡θ+4sin⁡θcos⁡θ+8sin⁡2θ=8 \cos ^{2} \theta+16 \sin \theta \cos \theta+4 \sin \theta \cos \theta+8 \sin ^{2} \theta =8+20sin⁡θcos⁡θ=8+20 \sin \theta \cos \theta

=8+10sin⁡2θ=8+10 \sin 2 \theta

Max Area =8+10=18(sin⁡2θ=1)⇒θ=45∘=8+10=18(\sin 2 \theta=1) \Rightarrow \theta=45^{\circ}

(a+b)2=(4cos⁡θ+2sin⁡θ+2cos⁡θ+4sin⁡θ)2(a+b)^{2}=(4 \cos \theta+2 \sin \theta+2 \cos \theta+4 \sin \theta)^{2}

=(6cos⁡θ+6sin⁡θ)2=(6 \cos \theta+6 \sin \theta)^{2}

=36(sin⁡θ+cos⁡θ)2=36(\sin \theta+\cos \theta)^{2}

=36(2)2=36(\sqrt{2})^{2}

=72=72

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
5 April, Shift 1
Subject
Mathematics
Chapter
Application of Derivatives
Topic
Physical Problem On extremum.

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