Question

Use Lagrange multipliers to give an alternate solution to the indicated exercise in Section\({\bf{11}}{\bf{.7}}\)Exercise\(\;{\bf{32}}\).

Short answer

So, the distance is\(\frac{{5\sqrt {14} }}{{14}}\).

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 \((0,1,1)\)is given by:

\(d = \sqrt {{{(x - 0)}^2} + {{(y - 1)}^2} + {{(z - 1)}^2}} \)

\(\begin{array}{c}{d^2} = {x^2} + {(y - 1)^2} + {(z - 1)^2}\\ = f(x,y,z)\end{array}\)

Constraints\(g = x - 2y + 3z\,and\,k = 6\)

Use Lagrange multiplier

Using Lagrange multipliers, and solve for equations:

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

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

Comparing both sides we have:

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

Solve equation

Solve equation (1),(2) & (3) for\(x,y,z\)we have:

\(\begin{array}{l}x = \frac{\lambda }{2}\,\,\,\,....(5)\\y = \frac{{2 - 2\lambda }}{2}\,\,\,....(6)\\z = \frac{{2 + 3\lambda }}{2}\,\,\,\,...(7)\end{array}\)

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

\(\begin{array}{c}\frac{\lambda }{2} - 2\left( {\frac{{2 - 2\lambda }}{2}} \right) + 3\left( {\frac{{2 + 3\lambda }}{2}} \right) = 6\\7\lambda + 1 = 6\\\lambda = \frac{5}{7}\end{array}\)

Find value of variable

Substitute\(\lambda = \frac{5}{7}\)in equation (5), (6) & (7).

\(\begin{array}{l}x = \frac{5}{{14}}\\y = \frac{2}{7}\\z = \frac{{29}}{{14}}\end{array}\)

Compute distance

Compute\(d\)using these values.

\(\begin{array}{l}d = \sqrt {{{\left( {\frac{5}{{14}}} \right)}^2} + {{\left( {\frac{2}{7} - 1} \right)}^2} + {{\left( {\frac{{29}}{{14}} - 1} \right)}^2}} \\d = \sqrt {\frac{{25}}{{196}} + \frac{{25}}{{49}} + \frac{{225}}{{196}}} \\d = \frac{{5\sqrt {14} }}{{14}}\end{array}\)