How do you find two numbers whose difference is 100 and whose product is a maximum?

Question

Aurora Alvarado · Accepted Answer

Unfortunately, there is no solution for this question since the product can be made as big as you wish.  For any #x# and #y# that satisfy the condition,  #(x+1)# and #(y+1)# will yield a larger product assuming that #x# and #y# are nonnegative. #(x+1)(y+1)=xy+x+y+1>xy# I have a feeling that the question might have been either "whose sum is 100" or "whose product is minimum ."

Paisley Johnson · Answer

undefined To find two numbers whose difference is 100 and whose product is a maximum, we can use the concept of optimization. Let x be one of the numbers. Then, the other number would be x + 100, as their difference is 100.

The product of these two numbers, denoted by P(x), is given by the equation P(x) = x(x + 100).

To find the maximum product, we can take the derivative of P(x) with respect to x and set it equal to zero to find the critical points.

P'(x) = 2x + 100 = 0
=> 2x = -100
=> x = -50

Since the second derivative of P(x) is positive for all real numbers, the critical point x = -50 corresponds to a local minimum. Therefore, the maximum product occurs at the endpoints of the interval or at critical points.

Considering the endpoints of the interval, when x = 0, the other number is 100, and when x = 100, the other number is 200.

Evaluating the product at these points:
P(0) = 0(0 + 100) = 0
P(100) = 100(100 + 100) = 100 * 200 = 20,000

Hence, the maximum product is achieved when the two numbers are 100 and 200, with a product of 20,000.