Using Lagrange Multiploiers locate any extrema of the function
x^2 + y^2 + z^2
subject to x + z = 6 and y + z = 8
I got (4/3, 10/3, 14/3)
This problem can be transformed into a 1-D unconstrained optimisation problem in the variable z. This transformed problem has a unique minimum at z=14/3, which then gives the x and y for the minimum by substitution into the constraints.
As Captain Black says, you can use the two constraints to reduce this to a quadratic function of the single variable, z, then show that z= 14/3 is a minimum for that function and solve for x and y from the constraints.
But since this says to use "Lagrange Multipliers", we have where and are the multipliers.
That gives the three equations , , and . Since , that is the same as 2z= 2x+ 2y or z= x+ y. Then x= z- y and the first constraint is 2z- y= 6 while the second is y+ z= 8. Adding those equations 3z= 14 so z= 14/3. Then y= 8- 14/3= 10/3 and x= 4/3 .