Use Lagrange multipliers to find the max and min of the function $f(x,y)=xe^y$ subject to the constraint $x^2+y^2=6$.
My attempt:
I tried to find both partials and set them equal to $\lambda$ times the partial of the constraint and got the following equations
$$e^y=2x(\lambda)$$
$$x(e^y)=2y(\lambda) $$
I then solved for $x$ and $y$ and got :
$$x=\frac{e^y}{2\lambda}$$
$$y=\frac{xe^y}{2\lambda}$$
I then plugged those into the original equation and got
$$\frac{(x^2+1)(e^(2y) )}{4\lambda^2}=6$$
I'm confused where to go from here and any help will be greatly appreciated
Best Answer
We have, $$e^y = 2 \lambda x$$ $$x e^y = 2 \lambda y$$ thus, $$x = \frac{2 \lambda y}{e^y} = \frac{2 \lambda y}{2 \lambda x} = \frac{y}{x}$$ finally, $$x^2 = y$$ Now we replace into the original (constraint) equation $$y^2 + y - 6 = 0$$ You have now the constraint in the form of a quadratic equation, you can easily find the value of $y$ through solving the equation.