1. ## Help on a word problem please!

I am having a lot of trouble trying to solve this problem. I don't really know how to start it off. Any kind of help would be great!! Thanks a lot!

Let S denote the set of points with coordinates (x,y) such that 0 ≤ x ≤ 15 and y ^2 = 150 - 10x. Find the coordinates of the point(s) of S nearest to the origin. Also find the coordinates of the point(s) of S farthest from the origin. Justify your answer.

2. Originally Posted by kaiba4192
I am having a lot of trouble trying to solve this problem. I don't really know how to start it off. Any kind of help would be great!! Thanks a lot!

Let S denote the set of points with coordinates (x,y) such that 0 ≤ x ≤ 15 and y ^2 = 150 - 10x. Find the coordinates of the point(s) of S nearest to the origin. Also find the coordinates of the point(s) of S farthest from the origin. Justify your answer.
We want to minimize and maximize the function $d(y)=\frac{1}{100}(y^4-200y^2+22500)$ where $d(y)$ is the distance from the origin to $(\frac{150-y^2}{10},y)$ squared.

The reason we can minimize the distance squared is because $x^2$ is a monotonically increasing function on $[0,\infty)$.

Let's solve $d\;'(y) = 0$.
$4y^3-400y=0 \implies y(y+10)(y-10)=0\implies y=-10,0,10$.

$d(\pm10) = 125$ and $d(0) = 225$.

This means $y=-10,10$ is a local minimum and $y=0$ is a local maximum.

Now to check the endpoints. $0\leq x\leq 15 \implies -5\sqrt{6}\leq y \leq 5\sqrt{6}$.

$d(\pm5\sqrt{6}) = 150$.

Therefore $(5, 10)$ and $(5, -10)$ are the points in $S$ that minimize the distance to the origin.

Also the point $(15, 0)$ maximizes our desired distance.

3. thanks a lot for replying.

How did you you get the function $d(y)=\frac{1}{100}(y^4-200y^2+22500)$

4. Originally Posted by kaiba4192

How did you you get the function $d(y)=\frac{1}{100}(y^4-200y^2+22500)$
Using the distance formula, $D=\sqrt{\left(\frac{150-y^2}{10}\right)^2+y^2}$.
Then I let $d=D^2$.

5. ohh i see, but wouldn't letting $d=D^2$ change the result because we're not deriving the original function?

6. Originally Posted by chiph588@
The reason we can minimize the distance squared is because $x^2$ is a monotonically increasing function on $[0,\infty)$.
Since $D>0$, $D^2$ is minimized/maximized $\iff D$ is minimized/maximized.

7. ohh ok, i get it now, thanks a lot!!