1. ## Shortest line segment

Hello,
Author wants me to find the length of shortest line segment from positive x-axis to positive y-axis going through a point (a,b) in the first quadrant. Answer provided to me $(a^{\frac23}+b^{\frac23})^{\frac32}$.

I have no clue to answer this question.

2. ## Re: Shortest line segment

It's pretty straight forward.

Use the fact that the slope is the same between (0,y) and (a,b) and (a,b) and (x,0).

That gets you y in terms of x.

x is now the variable you will minimize on.

write the squared length as a function of x and do the usual minimization routine.

that will get you the x that minimizes the segment squared length and thus also minimizes the length.

use that x to calculate the minimum squared length and thus the minimum length.

3. ## Re: Shortest line segment

This problem is mathematically equivalent to the following problem posted on another site:

A fence 8 ft. high (h) runs parallel to a building and is 15 feet (d) from it. Find the length (L) of the shortest ladder that will reach from the ground across the top of the fence and to the wall of the building.

Here is the solution I provided:

If we ignore all dimensions of the ladder except the length, this is equivalent to minimizing the sum of the squares of the intercepts of a line passing through the point $\displaystyle (d,h)$ in the first quadrant of the $\displaystyle xy$-plane. Let $\displaystyle a$ and $\displaystyle b$ be the $\displaystyle x$-intercept and $\displaystyle y$-intercept respectively if this line. Thus, the function we wish to minimize is (the objective function):

$\displaystyle f(a,b)=a^2+b^2$

Now, using the two-intercept form for a line, we find we must have (the constraint):

$\displaystyle \frac{d}{a}+\frac{h}{b}=1$

Using Lagrange multipliers, we find:

$\displaystyle 2a=\lambda\left(-\frac{d}{a^2} \right)$

$\displaystyle 2b=\lambda\left(-\frac{h}{b^2} \right)$

and this implies:

$\displaystyle b=a\left(\frac{h}{d} \right)^{\frac{1}{3}}$

Substituting for $\displaystyle b$ into the constraint, there results:

$\displaystyle \frac{d}{a}+\frac{h}{a\left(\frac{h}{d} \right)^{\frac{1}{3}}}=1$

$\displaystyle a=d^{\frac{1}{3}}\left(d^{\frac{2}{3}}+h^{\frac{2} {3}} \right)$

Hence, we have:

$\displaystyle b=h^{\frac{1}{3}}\left(d^{\frac{2}{3}}+h^{\frac{2} {3}} \right)$

and so we find:

$\displaystyle f_{\min}=f\left(d^{\frac{1}{3}}\left(d^{\frac{2}{3 }}+h^{\frac{2}{3}} \right),h^{\frac{1}{3}}\left(d^{\frac{2}{3}}+h^{ \frac{2}{3}} \right) \right)=\left(d^{\frac{2}{3}}+h^{\frac{2}{3}} \right)^3$

Now, we need to take the square root of this since the objective function is the square of the distance we actually wish to minimize. Let [/TEX]L[/TEX] be the length of the ladder, and we now have:

$\displaystyle L_{\min}=\left(d^{\frac{2}{3}}+h^{\frac{2}{3}} \right)^{\frac{3}{2}}$

4. ## Re: Shortest line segment

Originally Posted by romsek
It's pretty straight forward.

Use the fact that the slope is the same between (0,y) and (a,b) and (a,b) and (x,0).

That gets you y in terms of x.

x is now the variable you will minimize on.

write the squared length as a function of x and do the usual minimization routine.

that will get you the x that minimizes the segment squared length and thus also minimizes the length.

use that x to calculate the minimum squared length and thus the minimum length.
Hello,
Would you show me your work notes because I didn't understand your answer. Have you used the same method as mentioned in post#3 to this thread?

5. ## Re: Shortest line segment

Originally Posted by Vinod
Hello,
Would you show me your work notes because I didn't understand your answer. Have you used the same method as mentioned in post#3 to this thread?
No because you never show any work.

Just copy MarkFL's answer. You clearly don't care if you ever actually learn the stuff or not.

6. ## Re: Shortest line segment

Originally Posted by romsek
No because you never show any work.

Just copy MarkFL's answer. You clearly don't care if you ever actually learn the stuff or not.
Hello,
We have to minimise $x^2+y^2$. The slope of line going through the point (0,y),(a,b) and (x,0) is $\frac{a}{x} +\frac{b}{y}=1$. So $y= \frac{bx}{x-a}$. Now we will differentiate $\sqrt{x^2+\frac{b^2x^2}{(x-a)^2}}$

7. ## Re: Shortest line segment

Originally Posted by Vinod
Hello,
We have to minimise $x^2+y^2$. The slope of line going through the point (0,y),(a,b) and (x,0) is $\frac{a}{x} +\frac{b}{y}=1$. Now how to get y in terms of x? Is there any different method other than Lagrange multipliers method?
smh....

$\dfrac a x + \dfrac b y = 1$

$\dfrac b y = 1 - \dfrac a x = \dfrac{x-a}{x}$

$\dfrac y b = \dfrac{x}{x-a}$

$y = \dfrac{b x}{(x-a)}$

simple algebra

Now let

$ll = x^2 + y^2 = x^2+ \dfrac{b^2 x^2}{(a-x)^2}$

Now as usual with extrema problems we differentiate what we are trying to minimize and set that derivative equal to $0$

$\dfrac{d~ll}{dx} = 2 x \left(\dfrac{a b^2}{(a-x)^3}+1\right)$

and

$\dfrac{d~ll}{dx} = 0 \Rightarrow x = a + (ab^2)^{1/3}$

This is the only real nontrivial solution

using this value of $x$ and after some algebra we obtain

$ll = \left(a^{2/3}+b^{2/3}\right)^3$

and as this is the squared length

$\ell = \sqrt{ll} = \left(a^{2/3}+b^{2/3}\right)^{3/2}$

Technically we should ensure that this is in fact a minimum using the 2nd derivative test. I leave that to you.