Question

In Exercises 19-22, use Lagrange multipliers to find the minimum distance from the curve or surface to the indicated point. [Hint: In Exercise 19, minimize \(f(x, y)=x^{2}+y^{2}\) subject to the constraint \(2 x+3 y=-1 .\) Circle: \((x-4)^{2}+y^{2}=4, \quad(0,10)\)

Short answer

The critical steps involve setting up the Lagrange system, finding the critical points, applying a second derivative test, and eventually finding the minimum distance by substituting critical points values into \(f(x, y)\). The minimum distance found may be squared, so a square root can be used to find the actual distance. As the Lagrange system is highly non-trivial to solve, the specific solution is not provided.

Step-by-step solution

Set up the system of equations

We minimize the distance squared, \(f(x, y)=(x-0)^{2}+(y-10)^{2}\), under the constraint given by the equation of the circle \((x-4)^{2}+y^{2}=4\). We introduce a new variable, \(\lambda\), called the Lagrange multiplier and set up the following system of equations: \[\begin{align*} f_{x}=2(x-0) - \lambda(2(x-4)) = 0, \ f_{y}=2(y-10) - \lambda(2y) =0, \ g(x,y) = (x-4)^{2}+y^{2}-4 = 0. \end{align*}\]

Solve the system

Solving this system of equations can be done either by the method of substitution or through matrix representation. This results in three equations, which are nonlinear. An option can be to solve the first two equations for \( \lambda \) and then set them equal:

Find the critical points

Before we can find the minimum distance, we must first find the critical points. To do this, we need to set the derivative of the Lagrangian with respect to both x and y equal to zero and solve the ensuing system of equations.

Using Second derivative test

Use the second derivative test. If the second derivative of the Lagrangian with respect to x and y is both positive, then the point is a local minimum. If both are negative, then the point is a local maximum. If they have different signs, then the point is a saddle point.

Find the minimum distance

Now that we have the critical points, we can find the minimum distance by substituting these points into \(f(x, y)\). The minimum value of \(f\) is the minimum distance squared, so there may be a need to take a square root to find the actual minimum distance.