   Chapter 14.8, Problem 49E

Chapter
Section
Textbook Problem

(a) Find the maximum value of f ( x 1 ,   x 2 ,   .   .   .   ,   x n )   =   x 1   x 2   .   .   .     x n n f ( x 1 ,   x 2 ,   .   .   .   ,   x n )   =   x 1   x 2   .   .   .     x n n given that x1, x2, . . . , xn are positive numbers and x1 + x2 + . . . + xn = c, where c is a constant.(b) Deduce from part (a) that if x1, x2, . . . , xn are positive numbers, then x 1   x 2   .   .   .     x n n ≤ x 1   +   x 2   +   .   .   .   +   x n n 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?

(a)

To determine

To find: The maximum value of f(x1,x2,...,xn)=x1x2...xnn .

Explanation

Given:

The given function is f(x1,x2,...,xn)=x1x2...xnn where x1,x2,...,xn are positive numbers and x1+x2+...+xn=c , c is constant.

Calculation:

The given function is f(x1,x2,...,xn)=x1x2...xnn .

The Lagrange multipliers f(x1,x2,...,xn)=λg(x1,x2,...,xn) is computed as follows,

f(x1,x2,...,xn)=λg(x1,x2,...,xn)fx1,fx2,...,fxn=λgx1,gx2,...,gxnfx1(x1x2...xnn),fx2(x1x2...xnn),fxn(x1x2...xnn)=λgx1(x1+x2+...+xnc),gx2(x1+x2+...+xnc),gxn(x1+x2+...+xnc)[1n(x1x2...xn)1n1(x2x3...xn)],[1n(x1x2...xn)1n1(x1x3...xn)]...[1n(x1x2...xn)1n1(x1x2...xn1)]=λ1,1,...,1

Thus, the value of f(x1,x2,...,xn)=λg(x1,x2,...,xn) is {[1n(x1x2...xn)1n1(x2x3...xn)],[1n(x1x2...xn)1n1(x1x3...xn)]...[1n(x1x2...xn)1n1(x1x2...xn1)]}=λ1,1,...,1

The result,

{[1n(x1x2...xn)1n1(x2x3...xn)],[1n(x1x2...xn)1n1(x1x3...xn)]...[1n(x1x2

(b)

To determine

To show: Thevalues x1,x2,...,xn are positive numbers then prove x1x2...xnnx1+x2+...+xnn is equal under which situations.

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