Set up and evaluate the optimization problems. (Enter your answers as comma-separated lists.) Find two positive integers such that their sum is 14, and the sum of their squares is minimized. Find two positive integers such that their sum is 14, and the sum of their squares is maximized.

Answer and Step-by-step explanation:

Let x and y be two positive integers and their sum is 14:

X + y = 14

And the sum of square of this number is:

f = x2 + y2

 = x2+ (14 – x)2

Differentiate with respect to x, we get:

F’(x) = [ x2 + (14 – x)2]’ = 0

        2x + 2(14-x)(-1) = 0

        2x +( 28 – 2x)(-1) = 0

     2x – 28 +2x = 0

        2x + 2x = 28

         4x = 28

       X = 7

Hence, y = 14 – x = 14 -7 = 7

Now taking second derivative test:

F”(x) > 0

For x = y = 7,f reaches its maximum value:

(7)2 + (7)2 = 49 + 49

                   = 98

F at endpoints x Є [ 0, 14]

F(0) = 02 + (14 – 0)2

       =  196

F(14) = (14)2 + (14 – 14)2

  = 196

Hence the sum of squares of these numbers is minimum when x = y = 7

And maximum when numbers are 0 and 14.

answer: the 4 kilogram bag

step-by-step explanation: this is because if you divide the number of kilograms into the total it will tell you how much each individual kilogram would be worth. 2 kilos is 7 and change 3 would be around 6 ish and the 4 kilo would be 4.6 and change and the 5 kilos would be like 4.88

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)