Two sides of a triangle have lengths ‘a’ and ‘b’ and the angle between them is θθ.
Question:

Two sides of a triangle have lengths ‘a‘ and ‘b‘ and the angle between them is θθ. What value of θθ will maximize the area of the triangle? Find the maximum area of the triangle also.

Solution:

As, the area of the triangle, $A=\frac{1}{2} a b \sin \theta$

$\Rightarrow A(\theta)=\frac{1}{2} a b \sin \theta$

$\Rightarrow A^{\prime}(\theta)=\frac{1}{2} a b \cos \theta$

For maxima or minima, $A^{\prime}(\theta)=0$

$\Rightarrow \frac{1}{2} a b \cos \theta=0$

$\Rightarrow \cos \theta=0$

$\Rightarrow \theta=\frac{\pi}{2}$

Also, $A^{\prime \prime}(\theta)=-\frac{1}{2} a b \sin \theta$

or, $A^{\prime \prime}\left(\frac{\pi}{2}\right)=-\frac{1}{2} a b \sin \frac{\pi}{2}=-\frac{1}{2} a b<0$

i. e. $\theta=\frac{\pi}{2}$ is point of maxima

Now,

The maximum area of the triangle $=\frac{1}{2} a b \sin \frac{\pi}{2}=\frac{a b}{2}$