Question

Use Lagrange multipliers to give an alternate solution to the indicated exercise in Section\({\bf{11}}.{\bf{7}}\)Exercise\({\bf{31}}\).

Short answer

The minimum distance is\(\frac{2}{{\sqrt 3 }}\).

Step-by-step solution

Method of Lagrange multipliers

To find the maximum and minimum values of\(f(x,y,z)\)subject to the constraint\(g(x,y,z) = k\)(assuming that these extreme values exist and\(\nabla g \ne {\bf{0}}\)on the surface\(g(x,y,z) = k)\):

(a) Find all values of\(x,y,z\), and\(\lambda \)such that\(\nabla f(x,y,z) = \lambda \nabla g(x,y,z)g(x,y,z) = k\)

And

(b) Evaluate\(f\)at all the points\((x,y,z)\)that result from step (a). The largest of these values is the maximum value of\(f\); the smallest is the minimum value of\(f\).

Assumption

The distance between a generic point\((x,y,z)\)and the specific point\((2,0, - 3)\)is given by:

\(\begin{array}{c}d = \sqrt {{{(x - 2)}^2} + {{(y - 0)}^2} + {{(z - ( - 3))}^2}} \\{d^2} = {(x - 2)^2} + {y^2} + {(z + 3)^2}\\ = f(x,y,z)\end{array}\)

Constraints\(g(x,y,z) = x + y + z\,\,\& \,\,k = 1\)

Use Lagrange multiplier

Using Lagrange multipliers, and solve for equations:

\(\nabla f = \lambda \nabla g\)

\(\begin{array}{c}\nabla ({(x - 2)^2} + {(y)^2} + {(z + 3)^2} = \lambda \nabla (x + y + z)\\2(x - 2) + 2(y) + 2(z + 3) = \lambda (1 + 1 + 1)\end{array}\)

Comparing both sides we have:

\(\begin{array}{*{20}{r}}{2x - 4 = \lambda \,\,\,..(1)}\\{2y = \lambda \,\,\,\,...(2)}\\{2z + 6 = \lambda \,\,\,\,...(3)}\\{x + y + z = 1\,\,\,....(4)}\end{array}\)

Solve equation

Solve equation (1),(2) & (3) for\(x,y\,\& z\)respectively.

\(\begin{array}{l}x = \frac{{\lambda + 4}}{2}\,\,\,...(5)\\y = \frac{\lambda }{2}\,\,\,\,...(6)\\z = \frac{{\lambda - 6}}{2}\,\,\,...(7)\end{array}\)

Substitute value of\(x,y\,\& z\)in equation (4) we have:

\(\begin{array}{l}1 = \frac{{\lambda + 4}}{2} + \frac{\lambda }{2} + \frac{{\lambda - 6}}{2}\\1 = \frac{{3\lambda }}{2} - 1\\\lambda = \frac{4}{3}\end{array}\)

Find value of variable

Put\(\lambda = \frac{4}{3}\)in equation (5), (6) & (7)

\(\begin{array}{l}x = \frac{8}{3}\\y = \frac{2}{3}\\z = \frac{{ - 7}}{3}\end{array}\)

Compute distance

\(d\)Compute using these value to get minimum distance.

\(\begin{array}{l}d = \sqrt {{{\left( {\frac{8}{3} - 2} \right)}^2} + {{\left( {\frac{2}{3} - 0} \right)}^2} + {{\left( { - \frac{7}{3} - ( - 3)} \right)}^2}} \\d = \sqrt {\frac{4}{9} + \frac{4}{9} + \frac{4}{9}} \\d = \frac{2}{{\sqrt 3 }}\end{array}\)