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 \(2 x+3 y=-1 .\) Line: \(2 x+3 y=-1, \quad(0,0)\)

Short answer

The minimum distance from the curve \(2x+3y=-1\) to the point (0,0) is \(\sqrt{\frac{13}{49}} = \frac{\sqrt{13}}{7}\)

Step-by-step solution

Set up Lagrange's equation

To set up Lagrange's equation, we use the formula \(L(x, y, λ) = f(x, y) - λ(g(x, y) - c)\) where \(L\) is the Lagrange's function, \(f(x, y)\) is the function we want to minimize or maximize, \(g(x, y)\) is the constraint, and \(λ\) is the lagrange multiplier. In our case, \(f(x, y) = x^{2} + y^{2}\), \(g(x, y) = 2x + 3y\), and the constraint \(c = -1\). So our Lagrange's equation becomes \(L(x, y, λ) = x^{2} + y^{2} - λ(2x + 3y + 1)\)

Differentiate Lagrange's equation

The next step is to differentiate Lagrange's equation with respect to \(x\), \(y\), and \(λ\) and set them equal to zero to solve for \(x\), \(y\), and \(λ\). These equations will give: \[ \begin{align*} \frac{∂L}{∂x} &= 2x - 2λ = 0 \ \frac{∂L}{∂y} &= 2y - 3λ = 0 \ \frac{∂L}{∂λ} &= -2x - 3y - 1 = 0 \end{align*} \]

Solve the system of equations

Solve the system of three equations we found in Step 2 simultaneously to get the values of \(x\), \(y\), and \(λ\). Solving the equations gives \(x = λ\), \(y = \frac{3}{2}λ\), and \(-2λ - 3(\frac{3}{2}λ) - 1 = 0\). Solving the equations will give \(λ = -\frac{2}{7}\), \(x = -\frac{2}{7}\), and \(y = -\frac{3}{7}\)

Find the minimum distance

With the calculated \(x\) and \(y\), insert the values back into \(f(x, y) = x^{2} + y^{2}\) to find the minimum distance. Substituting the values into this equation gives \(-\frac{2}{7}^{2} + -\frac{3}{7}^{2} = \frac{13}{49}\)