Minimum point of x^2+y^2 given that x+y=10?

Answer 1

(x+y)^2 = 10^2

x^2+y^2 + 2xy = 100

So x^2 + y^2 = 100 - 2xy

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

So x^2 + y^2 = 100 - 2x(10-x)= 100 - 20x + 2x^2 = 2(x^2-10x+100)

Now the min of x^2 + y^2 is the same as the min of 2(x^2-10x+100) or 2 times the min of (x^2-10x+100)

Now just find the min of this quadratic and multiply that result by 2.

Answer 2

Substitue x = 10 -y into x^2 + y^2

(10-y)^2+y^2

2y^2 -20y+100

vertex = -b/2a so y= 5

then x = 5

Answer 3

your given is a linear equation x+y=10 simple enough. picture a line falling from point (10,0) to (10,0).

now consider your question minimum value and x² + y² just thinking openly at point (5,5) the combined squared values are 25 + 25 = 50. At (6,4) it's 36 + 16 = 52 we can assume that the values continue to rise (9,1) would be 81+1 = 82 so point (5,5) would be the minimum of the point values squared