Originally Posted by
HallsofIvy I see nothing wrong with the arithmetic.
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 $\displaystyle 2x\vec{i}+ 2y\vec{j}+ 2z\vec{k}= \lambda(\vec{i}+ \vec{k})+ \mu(\vec{j}+ \vec{k})$ where $\displaystyle \lambda$ and $\displaystyle \mu$ are the multipliers.
That gives the three equations $\displaystyle 2x= \lambda$, $\displaystyle 2y= \mu$, and $\displaystyle 2z= \lambda+ \mu$. Since $\displaystyle 2x+ 2y= \lambda+ \mu$, 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 .