# Lagrange Multipliers with Two Constraints Examples 1

Recall that if we want to find the extrema of the function $w = f(x, y, z)$ subject to the constraint equations $g(x, y, z) = C$ and $h(x, y, z) = D$ (provided that extrema exist and assuming that $\nabla g(x_0, y_0, z_0) \neq (0, 0, 0)$ and $\nabla h(x_0, y_0, z_0) \neq (0, 0, 0)$ where $(x_0, y_0, z_0)$ produces an extrema in $f$) then we ultimately need to solve the following system of equations for $x$, $y$ and $z$ with $\lambda$ and $\mu$ as the Lagrange multipliers for this system:

(1)

Let’s look at some examples of using the method of Lagrange multipliers to solve problems involving two constraints.

## Example 1

**Find the extreme values of the function $f(x, y, z) = 3x – y – 3z$ subject to the constraint equations $x + y – z = 0$ and $x^2 + 2z^2 = 1$.**

Let $g(x, y, z) = x + y – z = 0$ and let $h(x, y, z) = x^2 + 2z^2 = 1$. Then in computing the necessarily partial derivatives, we have that:

(2)

Now we immediately have from the second equation that $\lambda = -1$. Substituting this into the first and third equations and we have:

(3)

Note that from the equations above we have that $x \neq 0$ (otherwise this would contradict equation 1) and $z \neq 0$ (otherwise this would contradict equation 2) $]]. Thus we can freely divide by $x$ and $z$. In doing so, the first equation implies that $\mu = \frac{2}{x}$ and the second equation implies that $\mu = -\frac{1}{z}$ and so $\frac{2}{x} = – \frac{1}{z}$ which implies that $\frac{x}{2} = – z$ which implies that $z = -\frac{x}{2} (*)$. Plugging this into equation 4 and we have that:

(4)

Now for $x = \frac{\sqrt{2}}{\sqrt{3}}$, from $(*)$ we immediately have that $z = – \frac{\sqrt{2}}{2\sqrt{3}} = -\frac{1}{\sqrt{6}}$. For $x = -\frac{\sqrt{2}}{\sqrt{3}}$, from $(*)$ we have that $z = \frac{\sqrt{2}}{2\sqrt{3}} = \frac{1}{\sqrt{6}}$.

Now from the equation $x + y – z = 0$, we have that the corresponding points of interest are $\left ( \frac{\sqrt{2}}{\sqrt{3}} , -\frac{\sqrt{3}}{\sqrt{2}}, -\frac{1}{\sqrt{6}} \right )$ and $\left ( -\frac{\sqrt{2}}{\sqrt{3}} , \frac{\sqrt{3}}{\sqrt{2}}, \frac{1}{\sqrt{6}} \right )$. The first point yields a maximum of $2 \sqrt{6}$ while the second point yields a minimum of $-2 \sqrt{6}$.

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