Home»Math Guides»Lagrange Multipliers for Optimization and Finding Extreme Points (Example 3)
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.
\[\begin{array}{l}3 = 2\lambda x\\\lambda = \frac{3}{{2x}}\\1 = 2\lambda y\\\lambda = \frac{1}{{2y}}\end{array}\]
Set the \(\lambda \) values equal to each other.
\[\begin{array}{l}\frac{3}{{2x}} = \frac{1}{{2y}}\\6y = 2x\\3y = x\end{array}\]
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.