I'm taking an economics course this summer, and while multivariable calculus is not required (Calc 1/2 are), the professor will explain how to use a couple multivariable concepts.
The textbook introduces the Lagrange technique for F(x,y) with the constraint G(x,y) as $L\left(x,y\right)=F\left(x,y\right)+\lambda G\left(x,y,\lambda\right)$.
Then $\frac{\partial }{\partial x}\left(L\left(x,y,\lambda \right)\right)\:\&\:\frac{\partial \:}{\partial \:y}\left(L\left(x,y,\lambda \right)\right)\:\&\frac{\partial \:\:}{\partial \:\:\lambda \:}\left(L\left(x,y,\lambda \:\right)\right)$.
Then set all 3 partial derivatives equal to zero and solve the system of equations.
I tried an example from Khan, and I cannot solve the system of equations, which makes me think I made an error somewhere. Here is my work so far:
$F\left(x,y\right)=x^2y\:with\:constraint\:\:x^2+y^2=1$
$L\left(x,y,\lambda \right)=x^2y\:+\:\lambda \:\left(x^2+y^2-1\right)$
$\frac{\partial \:}{\partial \:x}\left(x^2y\:+\:\lambda \:\:\:\left(x^2+y^2-1\right)\right)=2yx+2λx$
$\frac{\partial \:}{\partial \:\:y}\left(x^2y\:+\:\lambda \:\left(x^2+y^2-1\right)\right)=x^2+2λy$
$\frac{\partial \:}{\partial \:\lambda \:\:}\left(x^2y\:+\:\lambda \:\left(x^2+y^2-1\right)\right)=x^2+y^2-1$
Now setting up a system of equations:
$2yx+2λx=0$
$x^2+2λy=0$
$x^2+y^2-1=0$
I don't believe this system of equations is solvable. Any help would be greatly appreciated!
EDIT: Following up on Paul's suggestion below: With x=0, I get $x=0\:\&\:y=+2\:OR\:-2$
With $=−,\:I\:get\:the\:following\:potential\:pairs:\:\left(\sqrt{\frac{2}{3}},-\sqrt{\frac{1}{3}}\right),\left(-\sqrt{\:\frac{2}{3}},-\sqrt{\:\frac{1}{3}}\right),\left(\sqrt{\:\frac{2}{3}},\sqrt{\:\frac{1}{3}}\right),\left(-\sqrt{\frac{2}{3}\:},\sqrt{\:\frac{1}{3}}\right)$
Now, the x=0 possibilites don't fit into my constraint. The only values that fit into my constraint are $\left(\sqrt{\frac{2}{3}},-\sqrt{\frac{1}{3}}\right),\left(-\sqrt{\:\frac{2}{3}},-\sqrt{\:\frac{1}{3}}\right),\left(\sqrt{\:\frac{2}{3}},\sqrt{\:\frac{1}{3}}\right),\left(-\sqrt{\frac{2}{3}\:},\sqrt{\:\frac{1}{3}}\right)$
At this point, I now plug these 4 potential pairs into the original function, and see which output is the highest?
Could someone confirm if my logic here is correct? Thanks!