Minimization

**littlejodo** · November 5th 2008, 11:39 AM

Yet another test prep question. I have the solution but I, sadly, do not really understand what is going on. It may be due to confusion over notation, I don't know. Anyway, I will write out the problem and solution and mark where I get lost so that maybe some nice person can help clarify it.

Find the points on the curve y = x^2 nearest to the point (0, 1).

Solution:
Minimize the distance from the point (0, 1) to (x, y), squared, to eliminate square
roots: d2(x, y) = x^2 + (y − 1)^2 . What is the 'd2'? and where did this equation come from? Is it right to think it is the distance formula modified?

Then the solution shows that the first derivative of the d2(x, y) = x^2 + (y − 1)^2 = - x / y-1 and the y' of the original curve equation is 2x. I am totally fine with that step. BUT the next step gets me...

Combining these, 2x = -x/y-1 so y = 1/2 and x = +/- sqrt(2) / 2
I'm sorry, but I have no idea where this came from. It's not over yet...

Using the substitution, we get d2(x) = x^2 +(x^2 −1)^2 = x^4 −x^2 +1.
Is this just subbing in the original equation of y = x^2 into the distance one?

Set the derivative equal to zero to get:
d2'(x) =4x^3 - 2x = 0 when x = 0 (the local maximum of
the distance from (0, 1) at the bottom of the parabola, since
d2''(x) = 12x^2 - 2 which is negative when x = 0, and when
4x^2 - 2 = 0, again where x = +/- sqrt(2)/ 2 where d''2 (x) = 4 so these are local (and global) minima. I am fine with the first and second derivatives here, it is the usage of them here as well as what x equals that still confuses me (from above x = +/- sqrt(2)/2 ).

And that was the answer... although I am not sure which part was actually the answer. Is the answer that the x points closest to (0,1) are 0, +/- sqrt(2)/ 2 , and 4?

Thank you for your patience and help!

**skeeter** · November 5th 2008, 02:06 PM

Find the points on the curve y = x^2 nearest to the point (0, 1).

here is the method they utilized w/o all the mumbo-jumbo ... hopefully you can follow what is going on.

distance between the two points $(x,x^2)$ and $(0,1)$ is

$d = \sqrt{(x-0)^2 + (x^2-1)^2}$

if one minimizes $(x-0)^2 + (x^2-1)^2$ , then it also minimizes the square root of that value.

let $z = (x-0)^2 + (x^2-1)^2$

$z = x^2 + (x^4 - 2x^2 + 1)$

$z = x^4 - x^2 + 1$

$\frac{dz}{dx} = 4x^3 - 2x$

set $\frac{dz}{dx} = 0$ ...

$4x^3 - 2x = 0$

$2x(2x^2 - 1) = 0$

three critical values ... $x = 0$ , $x = \pm \frac{1}{\sqrt{2}}$

$\frac{d^2z}{dx^2} = 12x^2 - 2$

use the 2nd derivative test ...

$x = 0$ makes $\frac{d^2z}{dx^2} < 0$

$x = 0$ is the location of a relative max

both $x = \pm \frac{1}{\sqrt{2}}$ make $\frac{d^2z}{dx^2} > 0$

both of these x-values minimize z

so, two points are the closest to $(0,1)$ (because of the symmetry of $y = x^2$ )

$\left(\frac{1}{\sqrt{2}} , \frac{1}{2}\right)$ and $\left(-\frac{1}{\sqrt{2}} , \frac{1}{2}\right)$

**littlejodo** · November 5th 2008, 02:13 PM

oh my gosh!! that was so much easier to understand!!!! why don't they just write it like that to begin with? Thank you!!

Thread: Minimization - need help understanding

LinkBack

Thread Tools

Display

Minimization - need help understanding

Similar Math Help Forum Discussions

Minimization problem

Maximization/minimization/whatever

Optimization: Minimization

minimization problem

Search Tags