Problem 35 - College Algebra and Calculus: An Applied Approach

Question

Find three positive numbers \(x, y\), and \(z\) that satisfy the given conditions. The sum is 30 and the sum of the squares is a minimum.

Step-by-Step Solution

Answer

The three positive numbers are \(x = 10\), \(y = 10\), and \(z = 10\)

Step 1: Write Down the Equations

We have two equations to work with, \(x + y + z = 30\) from the sum condition and \(f(x,y,z) = x^2 + y^2 + z^2\) for the minimum sum of the squares. We will try to minimize \(f(x,y,z)\).

Step 2: Conversion to a Function of One Variable

Rewrite the equation in terms of one variable, this simplifies the problem. From the sum condition, we have \(z = 30 - x - y\). Substituting \(z\) into \(f(x,y,z)\), we get a new function, \(g(x,y) = x^2 + y^2 + (30 - x - y)^2\).

Step 3: Find the partial derivatives and solve them

Find the partial derivatives of \(g(x,y)\) with respect to \(x\) and \(y\). Then, set these resulting equations equal to zero and solve them to get the critical points. If done correctly, one should find \(x = y = 10\).

Step 4: Substitute Back the Results

Substitute \(x = 10\) and \(y = 10\) into the sum condition \(x + y + z = 30\) to get \(z = 10\).

Step 5: Confirm that it minimize the function

To confirm that these values indeed minimize the function \(g(x,y)\) and don't maximize it or represent saddle points, we can use the Second Partial Derivative Test or reconsider the symmetry of \(g(x,y)\) in combination with its boundedness. As this function is symmetric and bounded from below by zero, and our critical point is a single point within the area of definition, it indeed has to be a global minimum. However, a rigorous proof would use the Second Partial Derivative Test in a formal manner.