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

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