In a $$\Delta ABC,\,\angle B = {90^ \circ }$$    and $$b+a=4.$$   The area of the triangle is the maximum when $$\angle C$$  is :

Solution :

$$\eqalign{ & b\cos \,\theta = a \cr & \therefore \,b\cos \,\theta + b = 4 \cr & {\text{or, }}b = \frac{4}{{1 + \cos \,\theta }} \cr & \therefore \,a = \frac{{4\cos \,\theta }}{{1 + \cos \,\theta }} \cr & \therefore {\text{ area}} = \Delta = \frac{1}{2}\,ba\sin \,\theta = \frac{1}{2}.\frac{4}{{1 + \cos \,\theta }}.\frac{{4\cos \,\theta }}{{1 + \cos \,\theta }}.\sin \,\theta = \frac{{4\sin \,2\theta }}{{{{\left( {1 + \cos \,\theta } \right)}^2}}} \cr & \therefore \,\frac{{d\Delta }}{{d\theta }} = 4.\frac{{2\cos \,2\theta {{\left( {1 + \cos \,\theta } \right)}^2} + \sin \,2\theta .2\left( {1 + \cos \,\theta } \right)\sin \,\theta }}{{{{\left( {1 + \cos \,\theta } \right)}^4}}} \cr & \therefore \,\frac{{d\Delta }}{{d\theta }} = 0\,\,\, \Rightarrow \cos \,2\theta .\left( {1 + \cos \,\theta } \right) + \sin \,2\theta .\sin \,\theta = 0 \cr & {\text{or, }}\cos \,2\theta + \cos \,\theta = 0 \cr & \therefore \,\cos \,2\theta = - \cos \,\theta = \cos \,\left( {\pi - \theta } \right){\text{ or }}\theta = \frac{\pi }{3} \cr} $$
$$\therefore \,\Delta $$  is maximum when $$\theta = \frac{\pi }{3}.$$   (It cannot be minimum, since in the given situation minimum is 0.)