How to solve this optimization by using Lagrangian?

Question

Let $f(x,y)=x(1+y).$
When is the value given by this function maximized subject to $x^2+y^2=1$ ?
How to do it with Lagrange multipliers?

Answer 1

The Lagrangian $L=x+xy+\lambda(1-x^2-y^2)$ has $L_x=1+y-2\lambda x,\,L_y=x-2\lambda y$, so$$L_x=L_y=0\iff(\lambda=x=0,\,y=-1)\lor(x=2\lambda/(4\lambda^2-1),\,y=1/(4\lambda^2-1)).$$The first solution gives $x^2+y^2=1,\,L=0$. The second gives$$1=x^2+y^2=(4\lambda^2+1)/(4\lambda^2-1)^2\stackrel{t:=4\lambda^2}{\implies}t(t-3)=0,\,L=\frac{8\lambda^3}{(4\lambda^2-1)^2}.$$The maximum $L$ occurs when$$t=3,\,\lambda=\sqrt{3}/2,\,x=\sqrt{3}/2,\,y=1/2,\,L=3\sqrt{3}/4.$$