Hello,
I am stuck with this question ,:
find the minimum value of$\displaystyle (x + 1)^2 + y^2$where $\displaystyle x + y$are real numbers with $\displaystyle x+y=1$ justify your answer
can someone help please?
$\displaystyle (x+1)^2 + y^2$ is the distance squared from the point (-1,0) to (x,y)
So the question is asking the square of the minimum distance from (-1,0) to any point on the line $\displaystyle x + y = 1$.
i.e. the square of the distance from the point (-1, 0) to the line $\displaystyle x + y = 1$
you should get the distance to be $\displaystyle \sqrt{2}$,
and minimum $\displaystyle (x+1)^2 + y^2 = \sqrt{2}^2 = 2$
Actually, your approach is easier.
$\displaystyle
(x+1)^2 + y^2 = (x+1)^2 + (1-x)^2
= (x^2 + 2x + 1) + (1 - 2x + x^2)
= 2x^2 + 2 = 2(x^2 + 1)
$
Now clearly $\displaystyle 2(x^2 + 1)$ is minimum when $\displaystyle x=0$, which gives 2.
If you want to know the distance to line approach (don't worry about it, if you haven't learned this):
Rearranging, the line equation is $\displaystyle y = -x + 1$
The slope of the perpendicular line has m = 1
A perpendicular line with m = 1 through point (-1, 0) has the equation $\displaystyle y = x + 1$
Intersection of the line and the perpendicular line is at (0,1)
So the minimum distance is from (-1,0) to (0,1) which is sqrt(2).
If you are confused on some step, draw a diagram and the graphs.