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Two Sides of a Triangle Have Lengths 'A' and 'B' and the Angle Between Them is θ What Value of θ Will Maximize the Area of the Triangle? Find the Maximum Area of the Triangle Also. - Mathematics

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Question

Two sides of a triangle have lengths 'a' and 'b' and the angle between them is \[\theta\]. What value of \[\theta\] will maximize the area of the triangle? Find the maximum area of the triangle also.  

Sum
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Solution

\[\text { As, the area of the triangle, A } = \frac{1}{2}ab\sin\theta\]

\[ \Rightarrow A\left( \theta \right) = \frac{1}{2}ab\sin\theta\]

\[ \Rightarrow A'\left( \theta \right) = \frac{1}{2}\text { ab }\cos\theta\]

\[\text { For maxima or minima, A}'\left( \theta \right) = 0\]

\[ \Rightarrow \frac{1}{2}ab\cos\theta = 0\]

\[ \Rightarrow \cos\theta = 0\]

\[ \Rightarrow \theta = \frac{\pi}{2}\]

\[\text { Also, A }''\left( \theta \right) = - \frac{1}{2}ab\sin\theta\]

\[\text { or,} A''\left( \frac{\pi}{2} \right) = - \frac{1}{2}ab\sin\frac{\pi}{2} = - \frac{1}{2}ab < 0\]

\[\text { i . e } . \theta = \frac{\pi}{2} \text { is point of maxima }\]

\[\text { Now }, \]

\[\text { The maximum area of the triangle } = \frac{1}{2}ab\sin\frac{\pi}{2} = \frac{ab}{2}\]

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Chapter 18: Maxima and Minima - Exercise 18.5 [Page 73]

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RD Sharma Mathematics [English] Class 12
Chapter 18 Maxima and Minima
Exercise 18.5 | Q 11 | Page 73

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