Problem 36 - College Algebra and Calculus: An Applied Approach

Question

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.

Step-by-Step Solution

Answer

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

Step 1: 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 \(S = x^2 + y^2 + z^2\).

Step 2: 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 \(z = 1 - x - y\).

Step 3: 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 \(S = 3x^2 - 2x + 1\).

Step 4: 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.

Step 5: Find \(y\), \(z\)

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