Question

\({\bf{1}} - {\bf{12}}\)Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint.

\(f\left( {{x_1},{x_2}, \ldots ,{x_n}} \right) = {x_1} + {x_2} + \cdots + {x_n};\,\,\,\,x_1^2 + x_2^2 + \cdots + x_n^2 = 1\)

Short answer

Maximum value is\(f\left( {\frac{1}{{\sqrt n }},\frac{1}{{\sqrt n }}, \ldots \ldots ,\frac{1}{{\sqrt n }}} \right) = \sqrt n \)and minimum value is\(f\left( { - \frac{1}{{\sqrt n }},\frac{{ - 1}}{{\sqrt n }}, \ldots \ldots \ldots ,\frac{{ - 1}}{{\sqrt n }}} \right) = - \sqrt n \).

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\(\nabla f(x,y,z) = \lambda \nabla g(x,y,z)g(x,y,z) = k\)

(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\(f\).

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

Given \(f\left( {{x_1},{x_2}, \ldots ,{x_n}} \right) = {x_1} + {x_2} + \cdots + {x_n};\,\,\,\,g(\left( {{x_1},{x_2}, \ldots ,{x_n}} \right):x_1^2 + x_2^2 + \cdots + x_n^2 - 1 = 0\)

By Lagrange's method of multipliers find all\({x_1},{x_2}, \ldots \ldots ,{x_n}\)and\(\lambda \)such that

\(\vec \nabla f\left( {{x_1},{x_2}, \ldots \ldots ,{x_n}} \right) = \lambda \vec \nabla g\left( {{x_1},{x_2}, \ldots \ldots \ldots ,{x_n}} \right)\)

And \(g\left( {{x_1},{x_2}, \ldots \ldots ,{x_n}} \right) = k\)

Comparing both sides:

\(\begin{array}{l}{f_{{x_1}}} = \lambda {g_{{x_1}}}\\{f_{{x_2}}} = \lambda {g_{{x_2}}}\\{f_{{x_n}}} = \lambda {g_{{x_n}}}\end{array}\)

Solve equation

\(\begin{array}{l}1 = 2\lambda {x_1}\,\,\,......(1)\\1 = 2\lambda {x_2} \cdots (2)\\1 = 2\lambda {x_n}\,\,.....(n)\\x_1^2 + x_2^2 + \ldots \ldots \ldots + x_n^2 = 1\,\,\,.....(n + 1)\end{array}\)

Solve equation

Now\(\lambda \ne 0\)and\({x_1},{x_2}, \ldots \ldots \ldots ,{x_n} \ne 0\)

Then we have\({x_1} = {x_2} = {x_3} \ldots \ldots \ldots = {x_n} = \frac{1}{{2\lambda }}\)

Using this in equation\((n + 1)\)

\(\frac{1}{{4{\lambda ^2}}} + \frac{1}{{4{\lambda ^2}}} + \ldots \ldots + \frac{1}{{4{\lambda ^2}}} = 1\)

\(\begin{array}{l}\frac{n}{4} = {\lambda ^2}\\\lambda = \pm \frac{{\sqrt n }}{2}\end{array}\)

Then\({x_1} = {x_2} = \ldots \ldots \ldots = {x_n} = \pm \frac{1}{{\sqrt n }}\)

Value at critical point

Then all the possible extreme points are\(\left( { \pm \frac{1}{{\sqrt n }}, \pm \frac{1}{{\sqrt n }} \ldots \ldots \ldots , \pm \frac{1}{{\sqrt n }}} \right)\)

Evaluate function on these extreme values to find the maximum value.

\(f\left( {\frac{1}{{\sqrt n }},\frac{1}{{\sqrt n }}, \ldots \ldots ,\frac{1}{{\sqrt n }}} \right) = \sqrt n \)

And the minimum value is:

\(f\left( { - \frac{1}{{\sqrt n }},\frac{{ - 1}}{{\sqrt n }}, \ldots \ldots \ldots ,\frac{{ - 1}}{{\sqrt n }}} \right) = - \sqrt n \)