Let B>0 . Show that the maximum of f(x1, …, xn)=x1 x2 ⋯xn subject to the constraints x1+⋯+xn=B a

Let B>0 . Show that the maximum of f(x1, …, xn)=x1 x2 ⋯xn...

Let $B>0 .$ Show that the maximum of $$f\left(x_{1}, \ldots, x_{n}\right)=x_{1} x_{2} \cdots x_{n}$$ subject to the constraints $x_{1}+\cdots+x_{n}=B$ and $x_{j} \geq 0$ for $j=$ $1, \ldots, n$ occurs for $x_{1}=\cdots=x_{n}=B / n .$ Use this to conclude that $$\left(a_{1} a_{2} \cdots a_{n}\right)^{1 / n} \leq \frac{a_{1}+\cdots+a_{n}}{n}$$ for all positive numbers $a_{1}, \ldots, a_{n}$

Let $B>0 .$ Show that the maximum of
$$f\left(x_{1}, \ldots, x_{n}\right)=x_{1} x_{2} \cdots x_{n}$$
subject to the constraints $x_{1}+\cdots+x_{n}=B$ and $x_{j} \geq 0$ for $j=$
$1, \ldots, n$ occurs for $x_{1}=\cdots=x_{n}=B / n .$ Use this to conclude that
$$\left(a_{1} a_{2} \cdots a_{n}\right)^{1 / n} \leq \frac{a_{1}+\cdots+a_{n}}{n}$$
for all positive numbers $a_{1}, \ldots, a_{n}$

Key Concepts

Lagrange Multipliers

A technique used in constrained optimization to locate the extrema of a function subject to one or more constraints. By introducing auxiliary variables (Lagrange multipliers), the problem is transformed into solving a system in which the gradients of the function and the constraints are aligned, thereby ensuring that the constraint is satisfied at the optimum.

Constrained Optimization

This is the process of optimizing (maximizing or minimizing) a function subject to certain constraints, such as equalities or inequalities. It involves finding the best solution under given limitations, and is a central concept in optimization theory and applied mathematics.

Arithmetic-Geometric Mean Inequality

This fundamental inequality states that for any set of positive numbers, the geometric mean is always less than or equal to the arithmetic mean, with equality if and only if all the numbers are equal. It is widely used in various fields of mathematics to compare multiplicative and additive quantities.

Transcript

00:03 We're asked to prove a version of the geometric mean, arithmetic mean, inequality.

00:19 So we're given a constant b greater than 0.

00:24 And we're asked to find a maximum with a function f equals x1, x2 through xn, subject to the constraints, the sum of x1 through xn equals b, and each of the xj must be greater than equal to zero.

00:38 Well, first of all, we notice that our constraints x1 plus xm, b, xj greater than equal to 0, j equal 1 through n, this defines a closed and bounded set in rn, and therefore, since f is continuous, as it's a polynomial, follows that f has extreme values on this set.

01:36 The maximum value for f does exist.

01:45 Now, minimum value, which is zero, occurs one of the coordinates of zero.

02:10 So any of the coordinates could be zero.

02:19 So we want to maximize our function.

02:23 F of x1, x2 through xn equals x1 times x2 up through xn, subject to the constraint.

02:33 G of x1 through xn equals the sum, x1 through xn.

02:39 X n equals and then our constant b and that each of the x -j is must be greater than or equal to 0 for j equal 1 through n so we'll use the legrongian multiplier view this the legronde equations are well the gradients of f is this is the vector x2 x3 up through x n and this is x1 x3 up through xn and this is x1 x3 up through x n and so on until we get x1 x2 multiply it all the way up to x n minus 1 and we have that the gradient of g is the vector one one one what is n ones and we have our lagrarynged condition that the gradient of f is equal to lambda times the gradient of g so putting all this together we get that x2 times x3 times up to xn is equal to lambda x1 times x3 all the way up to xn equals lambda and we continue in this way until we get x1 times x2 all the way up to xn minus 1 equals lambda so we get these n equations in n plus 1 unknowns so we have to consider lambda as well now let's solve for x1, x2, and xn using our constraints.

04:30 Well, our lebronge equations imply these equations.

04:37 This implies that x2 times x3 up through xn is equal to x1 times x2 up through xn minus 1.

04:46 Likewise that x1 times x3 up through xn is equal to x1 times x2 all the way out to xn minus 1.

04:55 We can do this for all of the equations except for the last one.

04:59 So we get x1 times x2 up 2, and then xn minus 2, xn, remove xn minus 1, is also equal to x1, x2, all the way up to xn minus 1...

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