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 θ">θθ. What value of θ">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 $=\frac{1}{2} a b \sin \frac{\pi}{2}=\frac{a b}{2}$

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