0

$f(x,y) = \sqrt{x^2 + y^2 -10x -10y +50} + \sqrt{x^2 - 4x + 20} + \sqrt{y^2 -14y + 74}$, find minimum of $f(x,y)$ and value of $x,y$ at which it is attained. I solve it by partial derivatives, is there a method like using vectors, graph or some other algebraic technique to solve such problem?

Hanul Jeon
  • 27,376
Paracetamol
  • 624
  • Please add your method to the question – tryst with freedom Aug 29 '21 at 11:38
  • Setting partial derivatives wrt to x and y to be equal to zero . And then solve for x and y out of it , lagrange also does same thing i think – Paracetamol Aug 29 '21 at 11:47
  • Oops you are right, I should have checked. "Also i think in coordinate system its like minimizing distance between three fixed points and a varaible pt wdy think?", the thing is the points at least how I got it is not fixed, the variable points determines the fixed point. – tryst with freedom Aug 30 '21 at 02:23
  • I see then ig is then some oly inequalities might help ? – Paracetamol Aug 30 '21 at 05:16

1 Answers1

0

Writing the function as $f(x,y) = \sqrt{(x-5)^{2}+(y-5)^{2}} + \sqrt{(x-2)^{2}+16} + \sqrt{(y-7)^{2}+25}$,

you can eliminate the contribution of the first addendum, placing: $x=5$ and $y=5$.

In fact, with these values you have:

$f(5,5)=5+\sqrt{29}$

which is the minimum for the function.

To reply to Paracetamol:

The values of $(x,y)$ that individually minimize the three radicals are:

$x=5$ and $y=5$,

$x=2$,

$y=7$.

There are four possible combinations (the $(x.y)$ pairs):

$(2,5)$, $(2,7)$, $(5,5)$ e $(5,7)$.

The pair that makes the function minimal is $(5.5)$.

giuseppe mancò
  • 1,671
  • 2
    How one shows that this is indeed the minimum , its not like removing a square root term always give a minimum isnt ? – Paracetamol Aug 30 '21 at 17:38
  • 1
    Why just those 4 combinations u take , is it really necessary that those numbers where any square tern goes to zero is always giving a minimum ? U can have some other (alpha,beta ) not among 2,5,7 – Paracetamol Aug 31 '21 at 06:32
  • 1
    Your method is flawed. The minimum of $\sqrt{x^2+1} + \sqrt{(x+1)^2 + 1}$ is not $x = 0$ nor $-1$. If you keep changing the constants, you achieve different minima every time. – Toby Mak Aug 31 '21 at 12:06