Let’s represent the two numbers as \(x\) and \(y\), where \(x+y=100\). We want to maximize the product \(P=xy\).

We can express one variable in terms of the other using the constraint \(x+y=100\), say \(y=100−x\). Now we have \(P=x(100−x)\).

To find the maximum of \(P\), we’ll take its derivative with respect to \(x\), set it equal to zero, and solve for \(x\). Then we’ll find the corresponding \(y\) value. Let’s do this in R:

# Define the function P(x) = x(100 - x)
P <- function(x) {
  return(x * (100 - x))
}

# Find the critical point by taking the derivative and solving for x
# P'(x) = 100 - 2x
# Set P'(x) = 0 and solve for x
x <- 100 / 2  # x = 50

# Find the corresponding y value
y <- 100 - x  # y = 50

# Calculate the maximum product
max_product <- P(x)

# Output the results
cat("The maximum product of two numbers with a sum of 100 is:", max_product, "\n")
## The maximum product of two numbers with a sum of 100 is: 2500
cat("The two numbers are:", x, "and", y, "\n")
## The two numbers are: 50 and 50