8ee.webp" data-orsrc="https://uploads-pracbee-ramanuj.s3.ap-south-1.amazonaws.com/pyq-question-images/batch-001/88be3dcbe8ee.webp" loading="lazy" style="max-width:100%;height:auto;display:block;margin:0 auto;max-height:40vh;vertical-align:baseline" alt="JEE Main 2024 (Online) 5th April Morning Shift Mathematics - Application of Derivatives Question 29 English Explanation">

$$\begin{aligned} & P D=4 \cos \theta \Rightarrow P S=4 \cos \theta+2 \sin \theta \\ & D S=2 \sin \theta \\ & A P=4 \sin \theta \\ & Q A=2 \cos \theta \Rightarrow P Q=2 \cos \theta+4 \sin \theta \\ & \Rightarrow \text { Area of } P Q R S=4(2 \cos \theta+\sin \theta)(\cos \theta+2 \sin \theta) \\ &=4\left[2 \cos ^2 \theta+2 \sin ^2 \theta+5 \sin \theta \cos \theta\right] \\ &=8+10 \sin 2 \theta \end{aligned}$$

Area is maximum when $$\sin 2 \theta=1 \Rightarrow \theta=45^{\circ}$$

$$\Rightarrow$$ Maximum area $$=8+10=18$$

$$\therefore P S=4 \times \frac{1}{\sqrt{2}}+2 \times \frac{1}{\sqrt{2}}=\frac{6}{\sqrt{2}}$$

$$\begin{aligned} & P Q=2 \times \frac{1}{\sqrt{2}}+4 \times \frac{1}{\sqrt{2}}=\frac{6}{\sqrt{2}} \\ \therefore \quad & (a+b)^2=\left(\frac{6}{\sqrt{2}}+\frac{6}{\sqrt{2}}\right)^2=\left(\frac{12}{\sqrt{2}}\right)=(6 \sqrt{2})^2=72 \end{aligned}$$

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JEE Main 2024 (Online) 5th April Morning Shift Mathematics - Application of Derivatives Question 29 Hindi Explanation

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