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Chapter 13: Problem 36

Find three positive numbers \(x, y\), and \(z\) that satisfy the given conditions. The sum is 1 and the sum of the squares is a minimum.

Short Answer

Expert verified
The three positive numbers that satisfy the given conditions are \(x = 1/3\), \(y = 1/3\), and \(z = 1/3\).

Step by step solution

01

Set up the equations

First, it is needed to set up the system of equations based on the information given in the problem. We have \(x + y + z = 1\), which is the condition for the sum of the three numbers. Our goal is to minimize the sum of squares \(S = x^2 + y^2 + z^2\).
02

Express \(y\), \(z\) in terms of \(x\)

To reduce this operation to a one variable problem, \(y\) and \(z\) can be expressed in terms of \(x\). From \(x + y + z = 1\), we get \(y = 1 - x - z\) and \(z = 1 - x - y\).
03

Substitute \(y\), \(z\) in S

Substitute \(y\) and \(z\) into the function for the sum of squares, \(S = x^2 + y^2 + z^2\). We get \(S = x^2 + (1-x-z)^2 + (1-x-(1-x-z))^2\), which simplifies to \(S = 3x^2 - 2x + 1\).
04

Find the minimum of \(S\)

To find the minimum of the function, differentiate \(S\) with respect to \(x\) and set the derivative equal to zero. The derivative \(S' = 6x - 2\), setting it to zero gives \(x = 1/3\). Secondly, use second derivative test to confirm it's a minimum. The second derivative \(S'' = 6 > 0\), thus \(x = 1/3\) gives a minimum.
05

Find \(y\), \(z\)

Substitute \(x = 1/3\) into the equations we got in Step 2 to find \(y\) and \(z\). Both return \(y = z = 1 - x - x = 1/3\).

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