lagrange multiplier (with variables x,y,z)

Question

I'm new to this topic, pls can I get hints on how to solve it: Find the point $(x,y,z)$ obeying $g(x,y,z)=2x+3y+z-12=0$ for which $f(x,y,z)=4x^2+y^2+z^2$ is minimum. Thanks in advance.

Answer 1

Use the method of Lagrange Multiplers, the equations are of the form $$\nabla f = \lambda\nabla g$$

In this example we have $\nabla f = \langle8x, 2y, 2z\rangle$, and $\nabla g = \langle2,3,1\rangle$.

Thus our equations are:

$$8x = 2\lambda \tag I$$

$$2y = 3\lambda \tag{II}$$

$$2z = \lambda \tag{III}$$

$$2x + 3y + z = 12\tag {IV}$$

Using $\text{III}$ we solve for $x$ and $y$ in terms of $z$ to get, from $\text{I}$: $8x = 2(2z) = 4z \implies x = \frac{z}{2} $, and from $\text{II}$ we have $2y = 3(2z) = 6z \implies y = 3z$.

Now we use the constraint equation to solve for $z$, then use that to find $x$ and $y$.

$$2(1/2)z + 3(3z) + z = 12 \implies z = \frac{12}{11}$$

$$x = \frac{6}{11} $$ $$y = \frac{36}{11} $$

Thus our point is $\left(\dfrac{6}{11},\dfrac{36}{11},\dfrac{12}{11}\right)$.