1. ## Lagrange multiplier method

For the function f(x,y) = 2x^3 + 3xy + 2y^3,
if the values of c and y are constrained to lie on the straight line y+x=1 use the lagrange multiplier method to find the value of (x,y) where f(x,y) has a turning point.

So far I have computed f sub x, f sub y and f sub lambda to produce 3 equations which imply that y=1-x,
6x^2 + 3 -3x + lambda =0 and
6x^2 - 9x + 6 + lamda = 0

at this point I get stuck

For the function f(x,y) = 2x^3 + 3xy + 2y^3,
if the values of c and y are constrained to lie on the straight line y+x=1 use the lagrange multiplier method to find the value of (x,y) where f(x,y) has a turning point.

So far I have computed f sub x, f sub y and f sub lambda to produce 3 equations which imply that y=1-x,
6x^2 + 3 -3x + lambda =0 and
6x^2 - 9x + 6 + lamda = 0

at this point I get stuck
I cannot follow what you've done at all.

$\displaystyle L = 2x^3 + 3xy + 2y^3 + \lambda (y + x - 1)$.

$\displaystyle \frac{\partial L}{\partial x} = 0 \Rightarrow 6x^2 + 3y + \lambda = 0$ .... (1)

$\displaystyle \frac{\partial L}{\partial y} = 0 \Rightarrow 6y^2 + 3x + \lambda = 0$ .... (2)

$\displaystyle \frac{\partial L}{\partial \lambda} = 0 \Rightarrow y + x = 1$ .... (3)

Note:

From (1) and (2) it follows that

$\displaystyle 6x^2 + 3y = 6y^2 + 3x \Rightarrow 2x^2 - x - 2y^2 + y = 0$ .... (4)

From (3) it follows that

$\displaystyle x = 1 -y$ .... (5)

Solve equations (4) and (5) simultaneously, test nature of extrema etc.

3. My apologies in my post I inputted the wrong equations corresponding to f sub x, etc.
I found the same 3 equations initially and have now found the value of (x,y) by solving the equations simultaneously, I have never seen or used the Lagrange multiplier before however and I do not know what the next steps are?! I have an example I am trying to follow but I cannot see what they have done after they have their (x,y) values.