f(x,y) = d² = x² + y²
g(x,y) = 3x²-4xy+6y² = 140
Then by my calculations,
grad(f) = 2xi + 2yj
grad(g) = (6x-4y)i + (-4x+12y)j
Then grad(f) = λgrad(g) produces these equations
2x = λ(6x-4y) (eq.1)
2y = λ(-4x+12y) (eq.2)
rearranging 1 gives:
λ = 2x/(6x-4y) (eq.3)
rearranging 2 gives:
λ = 2y/(-2x+6y) (eq.4)
then by cross multiplying,
-2x² + 6xy = 3xy - 2y²
taking all terms to left hand side,
-2x² +3xy + 2y² = 0
(-2x - y)(x - 2y)=0
therefore x = y/-2 or x = 2y
This is where I am stuck, now I have a relationship between x and y, what do I do next??
I'm fairly sure of my working to that point
Going on from your advice, say I taken the case x=y/-2 and sub into the constraint equation g(x) = 3x^2 - 4xy....
I return y=±4. Now I have a value of x and y to I find the corresponding lambda? I am not that familiar with lagrange multipliers and my textbook doesnt cover the topic greatly so if you could guide me through the method step by step it would be greatly appreciated!
Why do you need to calculate the value of lambda? Get the values of x and y and hence the points (x, y) that give the extremum. Then test the nature of these solutions.