[imtoghai]

Here's the problem:

Bogota, Columbia, New York City and Montreal, Canada are all situated along the same longitude. The distance from Montreal to Bogota is 1,230 miles. The distance from New York to Bogota is approximately thirty more than three timws the distance from New York to Montreal. Find the distance between each pair of cities.

EDIT
NEVERMIND, figured it out. (:
I was right after all LOL.
It was just my teacher's grading that confused me.

[azngirl1114]

Draw a line and have M for Montreal in the middle and B for Bogota on the right
Label the distance between M and B 1230 as given in the problem
Base on the given information, NY is closer to Montreal than Bogota so put NY on the left of the line
Distance from M to NY is x
and distance from NY to B is 3x+30
since it's NY M B
distance between M and B is the distance between NY and B minus distance between NY and M, set the number equal to the equation
solve for x

Edit: er..didn't see your nevermind until after i typed this out...lol. oh wells

[imtoghai]

I have a new question.

Well, basically, I don't understand the Fixed Perimeter/Maximum Area & Fixed Area/Minimum Perimeter stuff. How do you find the minimum perimeter with a fixed area? That's what I'm mostly having trouble on.

[1xviet]

^ What is the question? It's hard to visualize by what you wrote. I believe it involves a quadratic equation though.

[imtoghai]

Here are example questions:
What is the minimum perimeter of a rectangular region with an area of 900 square square feet?
What is the maximum area of a rectangular garden with a perimeter of 48 feet?

We would be given a rectangle with a fixed area of, for example, 121 square feet, then we would be asked to find the minimum perimeter of that.

Or, we would be given a rectangle with a fixed perimeter of, for example, 40 inches, then asked to find the maximum area of that.

[october-rain]

Here are example questions:
What is the minimum perimeter of a rectangular region with an area of 900 square square feet?
What is the maximum area of a rectangular garden with a perimeter of 48 feet?

1) primary equation
2x+2y = P

2) secondary equation
xy=900
y = 900/x

3) plug y value in so you're solving for only one variable

2x + 2y = P
plug in y=900/x into primary equation
you get:
2x + 2 (900/x) = P
simplify:
2x + 1800/x = P

4) differentiate it
you get 2 - 1800/x^2 = P'
simplify:
P' = (2x^2 - 1800)/x^2
set the equation to zero since you're finding the minumum
P' = (2x^2 - 1800)/x^2 = 0

5) try to solve for x by setting the numerator to zero [you can't do it to the denominator cus then you get 0]

2x^2 - 1800 = 0
2x^2 = 1800
x^2 = 900
x = 30, - 30 [square rooted it. -30 will not be the answer since dimensions cannot be negative]

since x= 30 y =30
plug the value back into the primary equation: 2x + 2y = P

2(30) + 2(30) = P
60 +60 = 120 <- answer to the problem

check:
you can check by plugging in the values in the secondary equation
xy = 900
(30)(30) = 900
900=900

the next problem can be solved the same way
hope it helps!