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:
8
Step-by-step explanation:
Answer:
I need a picture to solve...
Step-by-step explanation:
6r - 3t = 6 /*8
6r + 8t = - 16 /*3
48r - 24t = 48
48r + 24t = - 48
---------------------- +
96r = 0
r = 0
t = ( -16 + 6r)/8 = - 16/8 = - 2
