Problem 9 - Calculus Early Transcendentals

Question

What two nonnegative real numbers with a sum of 23 have the largest possible product?

Step-by-Step Solution

Answer

The two nonnegative real numbers are \(x = \frac{23}{2}\) and \(y = \frac{23}{2}\). The maximum possible product is \(\frac{529}{4}\).

Step 1: Define variables

Let \(x\) and \(y\) be the two nonnegative real numbers for which we want to find the largest product. We are given that the sum of these two numbers is 23, so we have the equation: \(x+y=23\).

Step 2: Express 'y' in terms of 'x'

Solve the equation \(x+y=23\) for \(y\) to express \(y\) in terms of \(x\). We get: \(y = 23 - x\).

Step 3: Formulate the product

Our goal is to maximize the product of \(x\) and \(y\). The product is: \(P(x) = x*(23-x)\).

Step 4: Find the first derivative of the product

Differentiate the product \(P(x)\) with respect to \(x\) to find its critical points. We have: \(P'(x)= \frac{d}{dx} x(23 - x) = 23 - 2x\).

Step 5: Set the first derivative equal to zero

Set the first derivative equal to zero to find the critical points: \(23 - 2x = 0 \Rightarrow x = \frac{23}{2}\).

Step 6: Determine the optimal values of the variables

From the previous step, we found that \(x = \frac{23}{2}\). Substitute this value into the equation for \(y\): \(y = 23 - x = 23 - \frac{23}{2} = \frac{23}{2}\). This shows that \(x = y = \frac{23}{2}\) is the optimal solution for maximizing the product.

Step 7: Find the largest possible product

Use the optimal values of \(x\) and \(y\) to find the largest possible product: \(P(x) = x*(23-x) = \frac{23}{2}* \frac{23}{2} = \frac{529}{4}\). The largest possible product of two nonnegative real numbers with a sum of 23 is \(\frac{529}{4}\).