Finding Area: Solving a Homework Problem

[bondgirl007]

I'd find it with Pythagoras.

So will it be:

(5)^2 + (3)^2 = AS^2
AS = 5.83

 

[dynamicsolo]

I'd find it with Pythagoras.

So will it be:

(5)^2 + (3)^2 = AS^2
AS = 5.83

The station isn't necessarily going to be at the midpoint, so you can't just use a '3' for that one leg CS (in fact, I'll bet money that leg won't be 3). Let's call that leg x. Then we have

5^2 + x^2 = AS^2 .

What would you write for BS^2?

 

[bondgirl007]

For BS^2, I'd write:

7^2 + (6-x)^2 = BS^2.

 

[bondgirl007]

After these formulas, I solved for AS and BS and this is what I got:

AS = 5+x
BS = 13-x

And then I substituted them into the distance formula:

d = AS + BS
d= 5+x + 13-x
da/dx = 0

I'm really confused now.

 

[dynamicsolo]

For BS^2, I'd write:

7^2 + (6-x)^2 = BS^2.

Good! Now, in principle, you would take the square root of each of these expressions to get AS and BS individually. The sum of these two lengths will be the total length of the road. You would take a derivative of that sum to work out the critical point of this function. That value of x would tell you where to put the station to minimize the length of road.

(I take back part of what I typed. You are only asked to find where to put the station, so you just need to find x and 6-x .)

 

[bondgirl007]

So will I differentiate this formula:

d = AS^2 + BS^2
or d = AS + BS

 

[bondgirl007]

I differentiated d = AS^2 + BS^2 like this

d = AS^2 + BS^2
d = 25 + x^2 + 49 + (6-x)^2
da/dx = 2x - 2(6-x)
0 = 4x-12
x = 3.

The correct answer according my textbook is different though. :S

 

[dynamicsolo]

The correct answer according my textbook is different though. :S

Sorry, I was thinking it would be possible to be spared having to take derivatives of the square roots, but it looks like that's unavoidable...

OK, so you want to take the square roots to get AS and BS and then minimize the function AS + BS. The algebra looks awful: you will have two terms with square roots in the denominators. But, since we want to know where this derivative is zero, we can add the two fractions and just look at when the numerator is zero.

You can put each term then on one side of the equation and square both sides to get rid of the radicals. While it now looks like you have polynomials of fourth degree, you will also find that a bunch of terms cancel out, so you really only end up with a quadratic polynomial to solve for zero. One of the solutions will be negative, so we can toss it out.
The other solution is 5/2 . Since that was the distance from C to S, the distance from D to S will be 6 - (5/2) = 7/2 km. , which matches the 21/6 you mentioned.

 

[bondgirl007]

I'm still not sure which one I should differentiate. If I do AS + BS then I get AS = 5+x and BS = 13-x. Won't the x cancel out when adding?

 

[dynamicsolo]

What you have to differentiate is

d/dx [ sqrt{25 + x^2} + sqrt{49 + (6-x)^2} ]

and set the result equal to zero. The derivative is not all that simple.

By the way, sqrt(25 + x^2) is not 5 + x and
sqrt(49 + (6-x)^2) is not 7+(6-x) .

 

[bondgirl007]

Thank you soo much! I finally got this question!

 

[dynamicsolo]

Thank you soo much! I finally got this question!

I'm glad to hear that. The algebra is a lot of work at first, so it's very gratifying when terms finally start to cancel out again...