### Multivariable Optimization using Lagrange Multipliers Example 3

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 $$f(x,y) = 3x + y$$ subject to the constraint $${x^2} + {y^2} = 10$$.

Start by finding the gradients, which is just a vector of all the partial derivatives.

$\nabla f = \left\langle {3,1} \right\rangle$

$\nabla g = \left\langle {2x,2y} \right\rangle$

$\left\langle {3,1} \right\rangle = \lambda \left\langle {2x,2y} \right\rangle$

And set the components equal, so you will get two equations.

$3 = 2\lambda x$

$\lambda = \frac{3}{{2x}}$

$1 = 2\lambda y$

$\lambda = \frac{1}{{2y}}$

Set the lambda values equal to each other.

$\frac{3}{{2x}} = \frac{1}{{2y}}$

$6y = 2x$

$3y = x$

Now substitute this equation into our constraint function.

${x^2} + {y^2} = 10$

${(3y)^2} + {y^2} = 10$

$10{y^2} = 10$

$y = \pm 1$

And substitute these values back into $$3y = x$$. Substituting y = -1 yields x = -3, and substituting y = 1 yields x = 3. That means that the extreme points are (-3,-1) and (3,1). Now substitute in these points into the objective function to find out whether the point is a max or min.

$f( - 3, - 1) = - 10$

$f(3,1) = 10$

So (-3,-1) is a min and (3,1) is a max.

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 Ex1

Lagrange Multipliers Optimization Ex2

Second Derivative Test Examples:

Second Derivative Test Optimization Ex1