# Lagrange Multipliers with One Constraint Examples 2

Recall from The Method of Lagrange Multipliers page that with the Method of Lagrange Multipliers that if we have a function $z = f(x, y)$ subject to the constraint $g(x, y) = C$, then if extreme values exist and $\nabla g (x, y) \neq (0, 0)$ on the level curve $g(x, y) = C$ then we can obtain the critical points subject to the constraint by solving the system of equations below and then comparing the values of $f$ at these critical points.

(1)

Furthermore, if we have a function $w = f(x, y, z)$ subject to the constraint $g(x, y, z) = C$, then if extreme values exist and $\nabla g(x, y, z) \neq (0,0,0)$ on the level surface $g(x, y, z) = C$, then we can find these extreme values by solving the system of equations below and then comparing the values of $f$ at these critical points.

(2)

We will now look at some more examples of solving problems regarding Lagrange multipliers.

## Example 1

**Find the maximum and minimum values of the function $f(x, y, z) = xyz$ subject to the constraint $x^2 + y^2 + z^2 = 6$ using Lagrange multipliers.**

Let $g(x, y, z) = x^2 + y^2 + z^2 = 6$. Computing the necessary partial derivatives and we have that:

(3)

Now multiply the first equation by $x$, the second equation by $y$, and the third equation by $z$ to get:

(4)

Thus we can eliminate the multiplier $\lambda$ immediately to get that $x^2 = y^2 = z^2$. Thus, substituting $y^2 = x^2$ and $z^2 = x^2$ into the fourth equation and we get that:

(5)

Thus the points of interest for this problem are $\left ( \pm \sqrt{2}, \pm \sqrt{2}, \pm \sqrt{2} \right )$. It’s not hard to see that the maximum of $f(x, y, z) = xyz$ is $2 \sqrt{2}$, and the minimum of $f(x, y, z)$ is $-2 \sqrt{2}$.

## Example 2

**Find the maximum and minimum values of the function $f(x, y, z) = xyz$ subject to the constraint $x^2 + 2y^2 + 3z^2 = 6$ using Lagrange multipliers.**

This problem is very similar to example 1, however, it is a tad more difficult. Let $g(x, y, z) = x^2 + 2y^2 + 3z^2 = 6$. In computing the necessary partial derivatives we get that:

(6)

Note that from these equations that $x \neq 0$, $y \neq 0$, and $z \neq 0$, for if one of $x$, $y$, or $z$ equals $0$, then it would follow that $x = y = z = 0$ which contradicts equation 4, $x^2 + 2y^2 + 3z^2 = 6$ since $0 \neq 6$! Thus we can divide by $x$, $y$, and $z$ freely.

From the first equation we have that $yz = \lambda (2x)$ implies that $\lambda = \frac{yz}{2x}$. Plugging this into the second equation and we have that:

(7)

The equation above implies that $x^2z = 2y^2z$ which implies that $x^2 = 2y^2 (*)$.

Now from the second equation we have that $\quad xz = \lambda (4y)$ implies that $\lambda = \frac{xz}{4y}$. Plugging this into the third equation and we have that:

(8)

The equation above implies that $2xy^2 = 3xz^2$ or equivalently, $2y^2 = 3z^2$. But from $(*)$ we have that then $3z^2 = x^2 (**)$.

Now plugging $(*)$ and $(**)$ into equation 4 and:

(9)

From $(*)$ we have that $2y^2 = x^2 = (\pm \sqrt{2})^2 = 2$, so $y^2 = 1$ and $y = \pm 1$. From $(**)$ we have that $3z^2 = x^2 = (\pm \sqrt{2})^2 = 2$, so $z^2 = \frac{2}{3}$ and $z = \pm \sqrt{\frac{2}{3}}$.

Thus our points of interest are $\left ( \pm \sqrt{2}, \pm 1, \pm \sqrt{\frac{2}{3}} \right )$. It’s not hard to see that a maximum is achieved when all coordinates are positive or only two coordinates are negative, and the maximum value is $\frac{2}{\sqrt{3}}$. Similarly, the minimum is achieved when all three coordinates are negative or only one coordinate is negative, and the minimum value is $- \frac{2}{\sqrt{3}}$.

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