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.

Short answer

The three positive numbers that satisfy the given conditions are \(x = 1/3\), \(y = 1/3\), and \(z = 1/3\). The minimum sum of their squares is \(1/3\).

Step-by-step solution

Set up the constraint

According to the problem conditions, the numbers \(x, y, z\) add up to 1. Formulate the constraint as: \(x + y + z = 1\)

Set up the function to minimize

We need to find the minimum of the sum of the squares, so we set the function as \(f = x^2 + y^2 + z^2\). However, with three variables it is a bit complicated to solve directly. Therefore, using the constraint from previous step to isolate one variable (let's pick \(z\)), we get: \(z = 1 - x - y\). Substitute this into the function to remove \(z\): \(f = x^2 + y^2 + (1 - x - y)^2 \).

Find partial derivatives

The next step involves finding the partial derivatives for the function in terms of both \(x\) and \(y\). The partial derivative with respect to \(x\) is: \(df/dx = 2x - 2(1 - x - y)\), and with respect to \(y\) is: \(df/dy = 2y - 2(1 - x - y)\).

Solve for x and y by setting partial derivatives to zero

Set the partial derivatives to zero and solve the resulting system of equations to find the potential optimal solutions. That gives: \(2x - 2(1 - x - y) = 0\) and \(2y - 2(1 - x - y) = 0\). From these, we find that \(x = y = 1/3\). Calculate \(z = 1 - x - y = 1/3\). This solution conforms to the condition that all numbers are positive and sums up to give 1. This candidate point is potentially the location of a minimum for the function.

Use the second derivative test to validate the minimum

The function is partial differentiable, so checking the second derivatives (or doing a second derivative test) should verify whether the found point is indeed a minimum. In this case, the second partial derivatives are \(d^2f/dx^2 = 2\), \(d^2f/dy^2 = 2\), which are both positive, indicating that (1/3, 1/3, 1/3) is indeed the point minimizing the function, i.e., the sum of squares. The minimum sum of squares is then \(f_{min} = (1/3)^2 + (1/3)^2 + (1/3)^2 = 1/3\).