Find the absolute maximum and minimum of $f(x,y,z)=ax+by+cz$ over the function $h(x,y,z)=x^2+y^2+z^2=1$ using Lagrange multiplier.
$$\mathcal{L}(x,y,z,\lambda)=f(x,y,z)+\lambda g(x,y,z)$$ $$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=ax+by+cz+\lambda (x^2+y^2+z^2-1)$$ $$\nabla_{x,y,z,\lambda} \mathcal{L}(x,y,z,\lambda)=(\frac{\partial \mathcal{L}}{\partial x},\frac{\partial \mathcal{L}}{\partial y},\frac{\partial \mathcal{L}}{\partial z},\frac{\partial \mathcal{L}}{\partial \lambda})$$ $$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=(a+2\lambda x,b+2\lambda y,c+2\lambda z,x^2+y^2+z^2-1)$$ And therefore:
$$\nabla_{x,y,z,\lambda} \mathcal{L}(x,y,z,\lambda)=0$$ $$\iff$$ $$a+2\lambda x=0$$ $$b+2\lambda y=0$$ $$c+2\lambda z=0$$ $$x^2+y^2+z^2-1=0$$
However I don't know how to continue,since solving the first three equations for the variables $x,y,z$ respectively does not let us to to solve the last equation just for a single variable $\lambda$,can someone helps me to finish the question?
You're on the right track. If you solve the first three equations for $x,y$, and $z$ and substitute them into the fourth equation, you can solve for $\lambda$. Once you know $\lambda$, you can determine $x,y$, and $z$ from the first, second, and third equation, respectively.