Question

(A) Find the maximum value of\(f\left( {{x_1},{x_2}, \ldots ,{x_n}} \right) = \sqrt(4){{{x_1},{x_2} \cdots {x_n}}}\)given that \({x_1},{x_2}, \ldots ,{x_n}\) are positive numbers and\({x_1} + {x_2} + \cdots + {x_n} = c\), where \(c\)is a constant.

(B) Deduce from part (a) that if \({x_1},{x_2}, \ldots ,{x_n}\) are positive numbers, then

This inequality says that the geometric mean of \(n\)numbers is no larger than the arithmetic mean of the numbers. Under what circumstances are these two means equal?

Short answer

1. The maximum value is\(\frac{c}{n}\).
2. \(\sqrt(n){{{x_1}{x_2} \ldots \ldots {x_n}}} = \frac{{{x_1} + {x_2} + \ldots \ldots + {x_n}}}{n}\)

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

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

(A)

Consider\(f\left( {{x_1},{x_2}, \ldots \ldots \ldots ,{x_n}} \right) = \sqrt(n){{{x_1},{x_2}, \ldots \ldots .{x_n}}}\)

Assume\(g\left( {{x_1},{x_2}, \ldots \ldots {x_n}} \right) = {x_1} + {x_n} + \ldots \ldots .. + {x_n} = c\)

Take \(h\left( {{x_1},{x_2}, \ldots \ldots {x_n}} \right) = {x_1},{x_2}, \ldots \ldots {x_n}\)

Apply Lagrange’s multiplier

By Lagrange's method of multipliers:

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

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

Differentiating & comparing we have :

\(\begin{array}{l}{x_2}{x_3} \ldots \ldots {x_n} = \lambda \,\,\,\,....(1)\\{x_1}{x_3} \ldots \ldots {x_n} = \lambda \,\,\,\,\,....(2)\\.\\.\\.\\.\\{x_1}{x_2} \ldots \ldots {x_{n - 1}} = \lambda \,\,\,\,\,\,...(n - 1)\\{x_1} + {x_2} + \ldots \ldots + {x_n} = c\,\,\,\,\,\, \cdots (n)\end{array}\)

Solve system

For \(c \ne 0\)from equation (1), (2)…..(n-1)

\(n\left( {{x_1}{x_2} \ldots \ldots {x_n}} \right) = \lambda \left( {{x_1} + {x_2} + \ldots \ldots .. + {x_n}} \right)\)

On using equation \(\left( n \right)\)

\(n\left( {{x_1}{x_2} \ldots \ldots \ldots {x_n}} \right) = \lambda c\)

i.e. \({x_1}{x_2} \ldots \ldots \ldots {x_n} = \frac{{\lambda c}}{n}\)

From equation (1)

\(\begin{array}{c}{x_1}{x_2} \ldots \ldots \ldots {x_n} = \lambda {x_1}\\\frac{{\lambda c}}{n} = \lambda {x_1}\\{x_1} = \frac{c}{n}\end{array}\)

Similarly \({x_2} = {x_3} = \ldots \ldots \ldots .. = {x_n} = \frac{c}{n}\)

Solve system for\({\bf{c = 0}}\)

If\(c = 0\), since \({x_1},{x_2}, \ldots \ldots \ldots ,{x_n}\) are positive numbers \(then\,\,{x_1} = {x_2} = \ldots \ldots \ldots = {x_n} = 0\)

Then the extreme points of \(h\) are\(\left( {\frac{c}{n},\frac{c}{n}, \ldots \ldots \ldots ,\frac{c}{n}} \right),(0,0, \ldots \ldots ..0)\)

Now

\(\begin{array}{c}h\left( {\frac{c}{n},\frac{c}{n}, \ldots \ldots .\frac{c}{n}} \right) = \left( {\frac{c}{n}} \right)\left( {\frac{c}{n}} \right) \ldots \ldots .\left( {\frac{c}{n}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {\frac{c}{n}} \right)^n}\\h(0,0, \ldots \ldots .,0) = 0\end{array}\)

Then the maximum value of \(h\) is \({\left( {\frac{c}{n}} \right)^n}\)

Now if \(h\) is maximum then\(f\) is also maximum, and then the maximum value of \(f\)is \({\left( {{{\left( {\frac{c}{n}} \right)}^n}} \right)^{1/n}} = \frac{c}{n}\).

Find geometric mean

(B)

From part\(({\rm{A}})\)we find that the maximum value of \(f\left( {{x_1},{x_2}, \ldots {x_n}} \right) = \sqrt(n){{{x_1}{x_2} \ldots \ldots {x_n}}}\,is\,\,\frac{c}{n}\)Where \(c = {x_1} + {x_2} + \ldots \ldots \ldots + {x_n}\)

That is for all \({x_1},{x_2}, \ldots \ldots \ldots ,{x_n}\)

\(\begin{array}{l}\sqrt(n){{{x_1}{x_2} \ldots \ldots {x_n}}} \le \frac{c}{n}\\\sqrt(n){{{x_1}{x_2} \ldots {x_n}}} \le \frac{{{x_1} + {x_2} + \ldots \ldots + {x_n}}}{n}\end{array}\)

 Step 7: Solve

But if \({x_1} = {x_2} = \ldots \ldots \ldots \ldots = {x_n}\)

Then \(\sqrt(n){{{x_1}{x_2} \ldots \ldots \ldots {x_n}}}\)

\(\begin{array}{l} = {\left( {x_1^n} \right)^{1/n}}\\ = {x_1}\end{array}\)

And \(\frac{{{x_1} + {x_2} + \ldots \ldots \ldots {x_n}}}{n}\)

\(\begin{array}{l}\frac{{{x_1} + {x_2} + \ldots \ldots \ldots {x_n}}}{n} = \frac{{n\left( {{x_1}} \right)}}{n}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {x_1}\end{array}\)

That is if \({x_1} = {x_2} = \ldots \ldots \ldots = {x_n}\) then\(\sqrt(n){{{x_1}{x_2} \ldots \ldots {x_n}}} = \frac{{{x_1} + {x_2} + \ldots \ldots + {x_n}}}{n}\).