Explanation Given : f ( x 1 , ..... x n ) = x 1 x 2 ... x n , x 1 + .... + x n = B , x j ≥ 0 Formula Used : ∇ f = λ ∇ g Proof : Step 1: Noticing that the constraints x 1 + .... + x n = B and x j ≥ 0 for j=1...n define a closed and bounded set in the nth dimensional space, hence f (continuous as a polynomial) has extreme values on this set. The minimum value zero occurs where one of the coordinates is zero (for example, for n=2, the constraint x 1 + x 2 = B , x 1 ≥ 0 , x 2 ≥ 0 is a triangle in the first quadrant) We need to maximize the function f ( x 1 , ... , x n ) = x 1 x 2 ... x n subject to the constraints g ( x 1 , ... , x n ) = x 1 + x 2 + ... + x n − B = 0 , x j ≥ 0 , j = 1 , .... , n . Step 2: While writing out the Lagrange Equations, the gradient vectors are as follows: ∇ f = 〈 x 2 x 3 ... x n , x 1 x 3 ... x n , x 1 x 2 ... x n − 1 〉 ∇ g = 〈 1 , 1.... , 1 〉 The Lagrange Condition ∇ f = λ ∇ g yields the following equations: x 2 x 3 .. x n = λ x 1 x 3 .. x n = λ x 1 x 2 .. x n − 1 = λ Step 3: Solving x 1 , x 2 .. x n using the constraint. The Lagrange equations imply the following equations: x 2 x 3 .. x n = x 1 x 2 .. x n − 1 x 1 x 3 .. x n = x 1 x 2 .. x n − 1 x 1 x 2 x 4 .. x n = x 1 x 2 ... x n − 1 ⋮ x 1 x 2 ... x n − 2 x n = x 1 x 2 ... x n − 1 We may assume that x j ≠ 0 for j=1, ...

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ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 15.8, Problem 53E

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To prove: The maximum of $f (x_{1}, ..... x_{n}) = x_{1} x_{2} ... x_{n}$ subject to the constraints $x_{1} + .... + x_{n} = B$ and $x_{j} \geq 0$ for j=1...n occurs for $x_{1} = .... = x_{n} = \frac{B}{n}$ and to show that

${(a_{1} a_{2} ... a_{n})}^{\frac{1}{n}} \leq \frac{a_{1} + ... + a_{n}}{n}$

for all positive numbers $a_{1}, ..., a_{n}$

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Chapter 15.8, Problem 52E

Chapter 15.8, Problem 54E

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Chapter 15 Solutions

CALCULUS (CLOTH)

Ch. 15.1 - Prob. 1PQ Ch. 15.1 - Prob. 2PQ Ch. 15.1 - Prob. 3PQ Ch. 15.1 - Prob. 4PQ Ch. 15.1 - Prob. 5PQ Ch. 15.1 - Prob. 1E Ch. 15.1 - Prob. 2E Ch. 15.1 - Prob. 3E Ch. 15.1 - Prob. 4E Ch. 15.1 - Prob. 5E

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To prove : The maximum of f ( x 1 , ..... x n ) = x 1 x 2 ... x n subject to the constraints x 1 + .... + x n = B and x j ≥ 0 for j=1...n occurs for x 1 = .... = x n = B n and to show that ( a 1 a 2 ... a n ) 1 n ≤ a 1 + ... + a n n for all positive numbers a 1 , ... , a n

Solution Summary: The author explains that f (continuous as a polynomial) has extreme values on this set. The minimum value zero occurs where one of the coordinates is zero.

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Chapter 15 Solutions

To prove : The maximum of f ( x 1 , ..... x n ) = x 1 x 2 ... x n subject to the constraints x 1 + .... + x n = B and x j ≥ 0 for j=1...n occurs for x 1 = .... = x n = B n and to show that ( a 1 a 2 ... a n ) 1 n ≤ a 1 + ... + a n n for all positive numbers a 1 , ... , a n

Solution Summary: The author explains that f (continuous as a polynomial) has extreme values on this set. The minimum value zero occurs where one of the coordinates is zero.

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To prove: The maximum of f(x1,.....xn)=x1x2...xn subject to the constraints x1+....+xn=B and xj≥0 for j=1...n occurs for x1=....=xn=Bn and to show that (a1a2...an)1n≤a1+...+ann for all positive numbers a1,...,an

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Chapter 15 Solutions