### Multivariable Optimization using Lagrange Multipliers Example 1

We can use Lagrange multipliers to solve problems where we're asked to find the max/min of an objective function, subject to equation constraints.

Find the extreme values of $$f(x,y,z) = {x^2} + {y^2} + {z^2}$$ subject to both constraints $$x - y = 1$$ and $${y^2} - {z^2} = 1$$.

For Lagrange multipliers involving 2 constraint equations, we can use:

$\nabla f = \lambda \nabla g + \mu \nabla h$

Where lambda and mu are unknown values. We can also set the components of the gradients equal individually, so we can obtain 3 equations, each one with the variables x, y and z.

$\frac{{\partial f}}{{\partial x}} = \lambda \frac{{\partial g}}{{\partial x}} + \mu \frac{{\partial h}}{{\partial x}}$

$\frac{{\partial f}}{{\partial y}} = \lambda \frac{{\partial g}}{{\partial y}} + \mu \frac{{\partial h}}{{\partial y}}$

$\frac{{\partial f}}{{\partial z}} = \lambda \frac{{\partial g}}{{\partial z}} + \mu \frac{{\partial h}}{{\partial z}}$

To complete these equations, we need to find all of the partial derivatives. I will arrange them into a matrix for neatness:

$\begin{array}{*{20}{c}}{\frac{{\partial f}}{{\partial x}} = 2x}&{\frac{{\partial g}}{{\partial x}} = 1}&{\frac{{\partial h}}{{\partial x}} = 0}\\{\frac{{\partial f}}{{\partial y}} = 2y}&{\frac{{\partial g}}{{\partial y}} = - 1}&{\frac{{\partial h}}{{\partial x}} = 2y}\\{\frac{{\partial f}}{{\partial z}} = 2z}&{\frac{{\partial g}}{{\partial z}} = 0}&{\frac{{\partial h}}{{\partial x}} - - 2z}\end{array}$

Plug these partial derivatives into the three equations.

$2x = \lambda$

$2y = - \lambda + 2\mu y$

$2z = - 2\mu z$

The equations simplify nicely. From the third equation here, you'll see z=0 or mu=-1. Sometimes you'll notice there are many ways to solve the series of equations; however, you'll notice that you might get imaginary numbers. That's not valid, and you'll have to go back and re-solve the equations to get an answer.

So let's choose mu=-1. You can use it to solve the other equations to get y=-lambda/4, then plug that into x-y=1 to get lambda=4/3, so x=2/3 and y=-1/3. BUT, when you plug it into $${y^2} - {z^2} = 1$$ you'll get z as an imaginary number! Normally imaginary numbers are not within the scope of your calculus course, so you need to go back and re-solve the equations! So choose z=0 instead at the earlier step a few paragraphs above!

When z=0, plug that into the constraints to find the values of the other variables.

${y^2} - {(0)^2} = 1$

$y = \pm 1$

And plug these y values back into the other constraint, you'll have two cases:

$x - (1) = 1$

$x = 2$

$x - ( - 1) = 1$

$x = 0$

So that means the extreme points are (2,1,0) and (0,-1,0). What are these extreme points? They're probably a max or min. Just plug them back into the original objective function, $$f(x,y,z) = {x^2} + {y^2} + {z^2}$$, and see which one is the max and which one is the min.

The max is $$f(2,1,0) = 5$$.

The min is $$f(0, - 1,0) = 1$$.

Please see our other worked-out examples of multivariable optimization using either Lagrange multipliers or the second derivative test.

Lagrange Multiplier Examples:

Lagrange Multipliers Optimization Ex2

Lagrange Multipliers Optimization Ex3

Second Derivative Test Examples:

Second Derivative Test Optimization Ex1