Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}A\mathrm{be}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{triangle}.\mathrm{Then}, \\ \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{triangle}=\frac{1}{2}ab\sin \theta \\ \frac{dA}{d\theta }=\frac{1}{2}ab\cos \theta \\ \mathrm{For}\mathrm{maximum}\mathrm{or}\mathrm{minimum}\mathrm{values}\mathrm{of}A,\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{dA}{d\theta }=0 \\ \Rightarrow \frac{1}{2}ab\cos \theta =0 \\ \Rightarrow \cos \theta =0 \\ \Rightarrow \theta =\frac{\pi }{2} \\ \mathrm{Now}, \\ \frac{d^{2}A}{d{\theta }^2}=-\frac{1}{2}ab\sin \theta \\ \Rightarrow \frac{d^{2}A}{d{\theta }^2}=-\frac{1}{2}ab<0 \\ \mathrm{So},\mathrm{area}\mathrm{is}\mathrm{maximum}\mathrm{when}\theta =\frac{\pi }{2}. \\ \therefore \mathrm{Maximum}\mathrm{area}=\frac{1}{2}ab\sin \frac{\pi }{2}=\frac{1}{2}ab \end{gathered}$$