1. ## minimum value help

Hello,
I am stuck with this question ,:

find the minimum value of $(x + 1)^2 + y^2$where $x + y$are real numbers with $x+y=1$ justify your answer

2. $(x+1)^2 + y^2$ 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 $x + y = 1$.
i.e. the square of the distance from the point (-1, 0) to the line $x + y = 1$
you should get the distance to be $\sqrt{2}$,
and minimum $(x+1)^2 + y^2 = \sqrt{2}^2 = 2$

I don't understand where the (-1,0) points appeared from....

But I done it this way, could someone tell me if this way is correct please?
$x+y=1,
y=1-x,
(x+1)^2 + y^2 ,= (x+1)^2 + (1-x)^2
=,x^2 +1 +1 -x^2$

= 2

4. What is the formula for the distance from (x,y) to (-1,0)?
sqrt((x+1)^2 + y^2) right? So the square of the distance is your formula.

(1 + x)^2 = 1 + 2x + x^2
(1 - x)^2 = 1 - 2x + x^2

5. could you show me how you got the distance to be the sqrt2 please

6. Actually, your approach is easier.

$
(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 $2(x^2 + 1)$ is minimum when $x=0$, 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 $y = -x + 1$
The slope of the perpendicular line has m = 1
A perpendicular line with m = 1 through point (-1, 0) has the equation $y = x + 1$
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.

7. Thanks

8. Originally Posted by wolfhound
Hello,
I am stuck with this question ,:

find the minimum value of $(x + 1)^2 + y^2$where $x + y$are real numbers with $x+y=1$ justify your answer

$(x+1)^2+y^2=(x+x+y)^2+y^2=(2x+y)^2+y^2=4x^2+4xy+2y ^2$
$=2\left[x^2+x^2+2xy+y^2\right]=2\left[x^2+(x+y)^2\right]=2\left[x^2+1\right]$