Problem 33 - 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 product is a maximum.

Step-by-Step Solution

Answer

The three positive numbers that satisfy the given conditions are \(x = 10\), \(y = 10\), and \(z = 10\).

Step 1: Formulate the Objective Function and Constraint

In this exercise, we want to maximize the product function \(P = xyz\). Given the constraint that the sum of the three numbers is 30, the constraint function is \(G(x, y, z) = x + y + z - 30 = 0\)

Step 2: Apply the Method of Lagrange Multipliers

The method of Lagrange multipliers involves setting the gradients of the objective function and the constraint function equal to each other, which gives us the following system of equations: \(\frac{dP}{dx} = \lambda \frac{dG}{dx}\), \(\frac{dP}{dy} = \lambda \frac{dG}{dy}\), and \(\frac{dP}{dz} = \lambda \frac{dG}{dz}\). After differentiating, these equations are \(y*z = \lambda\), \(x*z = \lambda\), and \(x*y = \lambda\). Given that \(x, y, z\) are nonzero, we can equate the equations that will give us \(y*z = x*z\) and \(x*y = y*z\). Therefore we have \(x = y = z\).

Step 3: Substitute into the Constraint

Now substituting \(x = y = z\) into the original constraint \(x + y + z = 30\) we get \(3x = 30\). Solving for \(x\) we find \(x = 10\). Therefore, since \(x = y = z\), each of the three numbers is \(10\).