Question

How to find the minimum value of : sqrt(4+y^2) + sqrt((x-2)^2+(2-y)^2) + sqrt((4-x)^2+1)

Answer 1

.

Answer 2

@RadEn looke on this please hope will be usefully


http://www.wikihow.com/Find-the-Maximum-or-Minimum-Value-of-a-Quadratic-Function-Easily

Answer 3

Can we do something like this?
1) Take derivative of it for both x, y
2) let numerator =0 to solve for y'
3) let y' =0 to solve for x
4) plug back to solve for y.
:) I might underestimate the problem.

Answer 4

u can solve either by use of calculus or can be solve it directly. its up to u.

Answer 5

how u wants to solve it?

Answer 6

Show me both, please

Answer 7

@Loser66 i know u can use calculus i solve second one method

Answer 8

Yes, please. I would like to learn more.

Answer 9

in first term if y is 0 then it give min value is sqrt 4 =2
in second term if x is 2 give (x-2) is 0 and if y is 2 then (2-y) is 0 so II term is 0 now
third one if x is 4 then x-4 is 0 and value is 1
so final min.value is 3
more explanation need ?

Answer 10

Using calculus, I got x=3, the same as your answer. I need know why you solve for radical =0 for all elements?

Answer 11

For the first one, \(\sqrt {4+y^2}\), it is >0 for all y, if y =-5, the result is still >0, but the minimum value of it is -5 , how can we know that just the radical =0, then that is the minimum one?

Answer 12

since we find min value and for it y must be 0 if any other number is taken it give more value than sqrt 4 so choose only y=0 give min value sqrt 4+y*y which is sqrt 4=2

Answer 13

I have to go, but still not understand it well. hehehe!! I am sorry for being stupid. Will be back later. Thanks for explanation.

Answer 14

http://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/min_max/min_max.html

Answer 15

the 3 terms look like distances:
the first is the distance from (0,0) to the point (2,y)
the middle is the distance from point (2,2)
the last is the distance from (4,0) to (x,1)

Graphically,

Answer 16

The intuition is point (2,1) minimizes the sum of the three terms.

Answer 17

in 3D, ie \(z = f(x,y)\), this is the addition of 2 perpendicular wedges/ valleys and a cone.

and, by my calc, \(f_x(2,y) \color{red}{= -\dfrac{2}{\sqrt{5}}} ~~~ \forall ~ y \ne 2 \) :( [there's a weird singularity at (2,2), which is the apex of the cone.]

the algebra associated with just setting \(f_x = f_y = 0\) suggests this is an exercise in finding an alternative clever solution.

can't see any ... moving the origin around just re-distributes the algebraic pain.

Answer 18

Check on wolfram, the answer is 5
http://www.wolframalpha.com/input/?i=minimize+(sqrt(4%2By%5E2)+%2B+sqrt((x-2)%5E2%2B(2-y)%5E2)+%2B+sqrt((4-x)%5E2%2B1))

Answer 19

so much for intuition!

The usual approach is to find the partial derivatives with respect to x and to y
and set both to zero. you get the pair of equations
\[ \frac{ x-2}{ \sqrt{ (x-2)^2+(y-2)^2} }+ \frac{x-4}{\sqrt{(x-4)^2+1}}=0 \\
\frac{ y-2}{ \sqrt{ (x-2)^2+(y-2)^2} }+ \frac{y}{\sqrt{y^2+4}}=0
\]

two equations and 2 unknowns, and x and y are not independent.
For both equations, move one of the terms to the other side, and square both sides
The first equation simplifies to
\[ (x-2)^2 = (x-4)^2 (y-2)^2 \]
or , after taking the square root:
\[ (x-2)= (x-4)(y-2) \text{ up to a sign}\]
based on the plot, we expect y to be less than 2, so (y-2) is negative
it's also reasonable to expect x<4, so (x-4) is negative.
that makes the right-hand side positive, and implies x > 2

Answer 20

using the same technique on the 2nd equation we will get
\[ (x-2)y= 2(y-2) \text{ up to a sign}\]
x>2 and we assume y>0 , so left side is positive
the right side is negative, so we will want to negate it to make the equation true. Thus the two equations simplify to
\[ (x-2)y= -2(y-2) \\ (x-2)= (x-4)(y-2) \]

Answer 21

at this point, it is a bit of messy (but doable) algebra to solve for x and y

Answer 22

using the 2nd equation
\[ (x-2)= (x-4)(y-2) \\ (y-2) = \frac{x-2}{x-4} \\ y= \frac{x-2}{x-4} +2\]
and replacing y-2 in the first equation we get
\[ (x-2)y= -2\frac{x-2}{x-4} \\ y= \frac{-2}{x-4}
\]

and replacing the y:
\[ \frac{x-2}{x-4} +2=\frac{-2}{x-4} \\ (x-2)+2x-8 = -2 \\ 3x= 8 \\ x= 8/3
\]
and
\[ y = \frac{-2}{\frac{8}{3}- \frac{12}{3}} = 2\cdot \frac{3}{4}= \frac{3}{2}\]

Answer 23

Got it. Thank you very much @phi. Your explanation can make me happy now, lol