Question

It's a quite long optimization question...

Consider 2 straight lines: y=1 and y=-1, and the point A(0,3) in the xy-plane. Take a point P on the straight line y=1 and a point Q on the line y=-1 such that PÄQ = 90°. Let the two points P and Q move preservering the above conditions. We are to find the minimum value of lenght of the line PQ.

First, denote the coordinates of P by (a,1) and Q by (b,-1). Then the condition PÄQ=90° is reduced to the conditions a!=0 and b!=0 and ab = (?, first question)

Since we know that a and b have opposite signs, let us assume that a<0 12 years ago

Answer 1

Then we have PQ² = (b-a)² + (?, second question)
= a²+b² + (?, third question)
\[\ge2|ab| + 9(?) = (?)\]\[PQ \ge?\]
Hence, when a = ? and b = ?, PQ takes the minimum value (?)

Answer 2

oh, and its 2|ab| + (?)

Answer 3

@Luigi0210 help him outt! :)

Answer 4

what can be the condition for the required length to be minimum?

Answer 5

I've realised 1 thing so far. P is in the 2nd quadrant and Q is in the 4st quadrant. I don't know how this will help me, though. Well, it's just a guess but the length of one of the sides must be maximized. I'm pretty bad at optimization problems

Answer 6

do you have the answer?

Answer 7

Yes, I do.

Answer 8

is it \(\large 2\sqrt2 \)
and the minimum length of PQ is 6 ???

Answer 9

Sec, gonna check.

Answer 10

Yes, the minimum value is 6

Answer 11

cool... that wasnt so bad....

Answer 12

a is -2sqrt(2), b is 2sqrt(2)
But I can't even start! gosh...

Answer 13

right... ok....

Answer 14

you'll need to find the relation between a and b...
you can do this with the pythagorean theorem AND distance formula...
just a sec for the drawing...

Answer 15

first, find:
d1 =
d2 =
???

Answer 16

Seems ok so far. I got here before but was stuck in the first question (a*b=?)

Answer 17

that's what we're gonna get to.... the relation between a and b...
after doing some substitutions, you should get
\(\large a \cdot b=-8 \) or \(\large a=\frac{-8}{b} \)

Answer 18

well, d1: \[\sqrt{(0-a)² +(3-1)²}\]
d2: \[\sqrt{(0-b)² +(3+1)²}\]

Answer 19

ok good... simplified, we get:

\(\large d_1=\sqrt{a^2+4} \) and \(\large d_2=\sqrt{b^2+16} \)

correct?

Answer 20

Yes. That's it.