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:
x²-6x+y²-4y=-4
Step-by-step explanation:
We know the circle equation stands like this:
(x-a)² + (y-b)² = R²
and (a,b) is center
and R stands for Radius
so in this case:
center = (3,2)
R = 3
So
(x-3)² + (y-2)² = 9
x²-6x+9 + y²-4y+4=9
x²-6x+y²-4y=-4
