Find two positive numbers whose product is 100 and whose sum is a minimum.

S=a+b

but a= 100/b
S=100/b + b

dS/db=0=-100/b^2 +1

solve for b, then go back and solve for a.

Exactly right. It's just the same as finding a rectangle of maximum area for a given perimeter. It's always a square.

i hope i got it right. The two numbers are 10 and 10?

thanks guys!

To find the two positive numbers whose product is 100 and whose sum is a minimum, we need to identify the values that satisfy this condition.

Let's assume the two positive numbers are x and y. According to the given conditions, we have:

1. The product of the two numbers is 100: xy = 100
2. The sum of the two numbers is a minimum: x + y = ?

To proceed, we will use the method of substitution:

1. Express one variable in terms of the other:
From the first equation, we can solve for x: x = 100/y

2. Substitute the expression for x into the second equation:
Substitute x = 100/y into x + y = ?
(100/y) + y = ?
Rearranging the terms, we get: (100 + y^2) / y = ?

Now, to minimize the sum (x + y) as y varies, we need to reduce the value of the fraction (100 + y^2) / y. To do this, we need to find the minimum value of the numerator (100 + y^2).

Since y is a positive number, y^2 is also positive, and as y increases, (100 + y^2) will increase. So, to find the minimum sum, we need to consider the value of y that makes (100 + y^2) as small as possible.

To find the minimum value of (100 + y^2), we can take its derivative with respect to y and set it equal to zero:

d(100 + y^2)/dy = 0
2y = 0
y = 0

At y = 0, the numerator (100 + y^2) is minimized. However, since we are looking for positive numbers, y = 0 is not a valid solution.

Therefore, there are no two positive numbers whose product is 100 and whose sum is a minimum.