Question

Use Lagrange multipliers to give an alternate solution the indicated exercise in Section\(\;{\bf{11}}.{\bf{7}}\)Exercise\({\bf{33}}\).

Short answer

Closet points are \((2,1,\sqrt 5 )\)and\((2,1, - \sqrt 5 )\).

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 the generic point\((x,y,z)\)and the specific point \((4,2,0)\)is given by:

\(d = \sqrt {{{(x - 4)}^2} + {{(y - 2)}^2} + {{(z - 0)}^2}} \)

\(\begin{array}{l}{d^2} = {(x - 4)^2} + {(y - 2)^2} + {z^2}\\{d^2} = f(x,y,z)\end{array}\)

Constraint is\(g(x,y,z):{x^2} + {y^2} - {z^2} = 0\)

Use Lagrange multiplier

Using Lagrange multipliers, and solve for equations:

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

\(\begin{array}{c}\nabla ({(x - 4)^2} + {(y - 2)^2} + {z^2}) = \lambda \nabla ({x^2} + {y^2} - {z^2})\\2(x - 4) + 2(y - 2) + 2(z - 2) = \lambda (2x + 2y - 2z)\end{array}\)

Comparing both sides we have:

Solve equation

Equation (3) gives\(z = 0\,or\,\lambda = - 1\).

For\(z = 0\).

Equation (4) gives\(x = 0\,and\,y = 0\).

However equation (1) is inconsistent with \(x = 0\)so \(z = 0\)does not produce any solutions. That means\(\lambda = - 1\)

Put \(\lambda = - 1\)in equation (1) to get\(x = 2\).

Equation (2) and \(\lambda = - 1\)give\(y = 1\).

Substituting the values for \(x\)and\(y\)into equation (4) gives \(z = \pm \sqrt 5 \).

The points on the cone closest to \((4,2,0)\)are \((2,1,\sqrt 5 )\)and\((2,1, - \sqrt 5 )\).