The product of the two positive numbers is 16. Let one of the numbers be X, the other number is `16/X` .

The sum of the two numbers is `S = X + 16/X` . The value of S is minimized for X such that `(dS)/(dX) = 0`

=> `1...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

The product of the two positive numbers is 16. Let one of the numbers be X, the other number is `16/X` .

The sum of the two numbers is `S = X + 16/X` . The value of S is minimized for X such that `(dS)/(dX) = 0`

=> `1 - 16/X^2 = 0`

=> `X^2 = 16`

=> X = 4

16/X = 4

The sum of one of them and the square of the other is `S2 = X^2 + 16/X` . S2 is minimized for the value of X where `(dS2)/(dX) = 0`

=> `2X - 16/X^2 = 0`

=> `X^3 = 8`

=> X = 2

16/X = 8

**The sum is the least for the set {4, 4} and the sum of one of the numbers and the square of the other is least for {2, 8}**

factors of 16: 1&16, 2&8, 4&4. `1+16=17` , `2+8=10` , `4+4=8` . For (a) the numbers are 4 and 4.

`1^2 +16=17` `16^2+1=257 `

`2^2 +8=12`` 8^2 +2 = 66`

`4^2 + 4=20 `