1. ## Extrema Problem

Show that there is a point on the line through (2,0,0) & (0,5,0) but that it is nearest to the line x=u, y=u, z=u.

I started by creating a function, F(x) = cx + d, where that f(c,d) = The sum from 1 to n of (cx_i + b - y_i)^(2) so I could minimize it. However, I just keep coming up blanks.

Show that there is a point on the line through (2,0,0) & (0,5,0) and that is nearest to the line x=u, y=u, z=u.

Let's see if I understood: the line through $\displaystyle (2,0,0)\,,\,(0,5,0)\,\mbox{ is }\,\{(2,0,0)+t(-2,5,0)\,\,\slash\,\,t\in\mathbb{R}\}=\{(2-2t,5t,0)\,\,\slash\,\,t\in\mathbb{R}\}$ , and apparently the other line is just $\displaystyle ((s,s,s)\,\,\slash\,\,s\in\mathbb{R}\}$, so take a general point in the first line and evaluate its distance from a general point of the second line (Apparently you don't know yet/you can't use , the well known formula for the distance of a given point from a given line):

$\displaystyle \sqrt{(2-2t-s)^2 +(5t-s)^2+s^2}=\sqrt{3s^2+29t^2-6st-4s-8t+4}$ , but of course the max.-min. points of the above are the same without the square root (why?), so define $\displaystyle f(s,t)=3s^2+29t^2-6st-4s-8t+4$ and show there's a unique solution to $\displaystyle f_s=f_t=0$ (I think it is $\displaystyle t = \frac{3}{13}$...)

Tonio

I started by creating a function, F(x) = cx + d, where that f(c,d) = The sum from 1 to n of (cx_i + b - y_i)^(2) so I could minimize it. However, I just keep coming up blanks.
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3. Tonio: Am I trying to "minimize" the function, f(s,t)?

Tonio: Am I trying to "minimize" the function, f(s,t)?

Well, more than trying to minimize it you're trying to prove it has a minimal point. Use the first and second order partial derivatives and then the Hessian of f.

Tonio

5. Originally Posted by tonio
Well, more than trying to minimize it you're trying to prove it has a minimal point. Use the first and second order partial derivatives and then the Hessian of f.

Tonio
Alright, I took the first and second partials and got,

fs(s,t) = 6s - 6t
ft(s,t) = 58t - 6s - 8

fss(s,t) = 6
fst(s,t) = 6
ftt(s,t) = 58
fts(s,t) = -6

Then, set up the Hessian, and got,

[6 6 ; 58 -6]

Where do I go from here?

Alright, I took the first and second partials and got,

fs(s,t) = 6s - 6t
ft(s,t) = 58t - 6s - 8

fss(s,t) = 6
fst(s,t) = 6
ftt(s,t) = 58
fts(s,t) = -6

Then, set up the Hessian, and got,

[6 6 ; 58 -6]

Where do I go from here?

That's wrong: $\displaystyle f_s=6s-6t\Longrightarrow f_{st}=-6=f_{ts}\,,\, f_{ss}=6\,,\,f_{tt}=58$ , so the Hessian is

$\displaystyle H_f=\left(\begin{array}{rr}6&-6\\-6&58\end{array}\right)$ , and this is a positive definite (symmetric, of course) matrix, so the critical point is a minimum one.

Tonio

7. Originally Posted by tonio
That's wrong: $\displaystyle f_s=6s-6t\Longrightarrow f_{st}=-6=f_{ts}\,,\, f_{ss}=6\,,\,f_{tt}=58$ , so the Hessian is

$\displaystyle H_f=\left(\begin{array}{rr}6&-6\\-6&58\end{array}\right)$ , and this is a positive definite (symmetric, of course) matrix, so the critical point is a minimum one.

Tonio
What is a positive definite matrix? And how did you determine the critical point was a min? Shouldn't I end up with a couple of ordered triple points as the solution to this problem?

In my book you should know this much linear algebra before you deal with multivariable functions and, at any rate, you should know how to classify critical points according to the function's Hessian or, at least, you should know what means $\displaystyle f_{ss}>0\,\,\,and\,\,\,f_{ss}f_{tt}-(f_{st})^2>0.$ , when the partial derivatives are evaluate at a critical point.