Two numbers have a difference of 20. The sum of their squares is a minimum. Determine the numbers.?

Question

This needs to be completed by completing the square. I know the answer is (10,-10), but I dont understand how to get it.

Answer 1

The average of the two numbers is 0.
The sum of their squares is 200.

Pick any two numbers which differ by 20 - let's say 11 and -9
11^2 + (-9)^2 = 202.

The other THREE posters used calculus which is uncalled for in this algebra problem. Always remember that neither of us will be available when you're taking your exam or test.

Answer 2

x-y=20 x=20+y
Sum= x^2+y^2= (20+y)^2+y^2= 2y^2+40y+400
Sum'=4y+40
set Sum'=0 so 4y+40=0 y=-10
check that this is at a minimum
Sum''=4= always concave up so -10 is at the minimum
x-y=20 x-(-10)=20 x=10

so the numbers are 10 and -10

Answer 3

Let x be one of the numbers
the other number will be x+20
y=sum of the squares
=x^2+[x+20]^2
=x^2+x^2+40x+400
=2[x^2+20x+200]
y'=2[2x+20]
we equte y' to zero to find the value of x
2[2x+20]=0
x=-10
the other number=z
z-x=20
z+10=20
z=10
The numbers are 10 and -10
This is based on the condition that
their sum of squares is minimum

Answer 4

Let the numbers be x and x-20.

y = x^2 + (x-20)^2 = x^2 +x^2 -40x +400
= 2x^2 -40x +400

dy/dx = 4x -40
0=4x -40, so x=10, x-20= -10.

Answer 5

x - y = 20
x^2 + y^2

x = y + 20

(y + 20)^2 + y^2
((y + 20)(y + 20)) + y^2
(y^2 + 20y + 20y + 400) + y^2
y^2 + 40y + 400 + y^2
2y^2 + 40y + 400
2(y^2 + 20y + 200)

y = (-b)/(2a)
y = (-20)/(2(1))
y = -10

x - y = 20
x - (-10) = 20
x + 10 = 20
x = 10

The values are (10,-10)

the y = (-b)/(2a), is to help find the maximum or minimum value of the parabola.

Answer 6

a - b = 20
a = 20 + b
a*a + b*b = z

(20+b)*(20+b) +b*b =z
400 +40b + 2(b*b) = z

The minimum of z is where the derivative,
db/dz, is 0

db/dz = 40 + 4b
0 = 40 + 4b

b = 10, a = -10