Question

$2$ sides of a triangle are given to find the angle between them such that area is maximum.

Answer 1

Area of $\Delta ABC=\left(A\right)=\frac{1}{2}\cdot AB\cdot AC\cdot \sin \theta$

$A=\frac{1}{2}\left(AB\right)\left(AC\right)\cdot \sin \theta$

$A=\frac{1}{2}\left(l_1\right)\left(l_2\right)\sin \theta$

$\frac{dA}{d\theta }=(\frac{1}{2}\cdot l_1\cdot l_2)\cdot \cos \theta$

If $\frac{dA}{d\theta }=0$

$(\frac{1}{2}\cdot l_1\cdot l_2)\cos \theta =0$

$\cos \theta =0$

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

$A_{max}=\frac{1}{2}\cdot l_1\cdot l_2\sin \frac{\pi }{2}$

$A_{max}=\frac{1}{2}\cdot l_1\cdot l_2$.