#### 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.

#### Solution

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

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

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

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

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

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

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

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

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

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

\[\text { Now }, \]

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