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pammywammy

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An Olympic track is constructed with a rectangle and half circles on each end. The Olympic track is 0.5 kilometers.

a) Draw a figure to represent the above problem. Let l and w represent the length and width of the rectangle, respectively.



b) Write the area A of the rectangular region as a function of I and in meters (not kilometers).



c) Write the area A of the rectangular regional as a function of w and in meters (not kilometers).
 
So, what have you done so far? Where are you stuck?

If you are waiting for someone here to do it for you .... not going to happen!
 
No i don’t want someone to do it with me. I am just stuck on b and c. I don’t know what I need to do for the área.
 
pammywammy said:
An Olympic track is constructed with a rectangle and half circles on each end. The Olympic track is 0.5 kilometers.
a) Draw a figure to represent the above problem. Let l and w represent the length and width of the rectangle, respectively.
b) Write the area A of the rectangular region as a function of I and in meters (not kilometers).
c) Write the area A of the rectangular regional as a function of w and in meters (not kilometers).
Click to expand...
A rectangle with length l and width w → what would be the expression of the area of the rectangle?

If you do not know the expression for area of a rectangle - look in your textbook or do a Google search and tell us what you found.
 
Subhotosh Khan said:
A rectangle with length l and width w → what would be the expression of the area of the rectangle?

If you do not know the expression for area of a rectangle - look in your textbook or do a Google search and tell us what you found.
Click to expand...
the expression of the area would be A= l x w
 
pammywammy said:
the expression of the area would be A= l x w
Click to expand...
Correct.

Now look at the figure you have drawn for part (a). It has two straight-edges and two semi-circular edges.

Look a the statement -
pammywammy said:
An Olympic track is constructed with a rectangle and half circles on each end. The Olympic track is 0.5 kilometers.
Click to expand...
What can you deduce from that?
 
Subhotosh Khan said:
Correct.

Now look at the figure you have drawn for part (a). It has two straight-edges and two semi-circular edges.

Look a the statement -
What can you deduce from that?
Click to expand...
so i came up with an equation to solve the are for w. i dont know if its right. pi x w +2x= 500 meters
 
pammywammy said:
i am confused
Click to expand...

the distance once around the track is 500 meters = 1/2 kilometer

the track consists os two rectangular lengths = [MATH]2L[/MATH]and one circular length (two halves) = [MATH]\pi W[/MATH]
therefore the total length in meters is given by the equation in my previous post

you need to solve the equation for L in terms of W to get A = LW in terms of W

you also need to solve the equation for W in terms of L to get A = LW in terms of L
 
skeeter said:
the distance once around the track is 500 meters = 1/2 kilometer

the track consists os two rectangular lengths = [MATH]2L[/MATH]and one circular length (two halves) = [MATH]\pi W[/MATH]
therefore the total length in meters is given by the equation in my previous post

you need to solve the equation for L in terms of W to get A = LW in terms of W

you also need to solve the equation for W in terms of L to get A = LW in terms of L
Click to expand...
skeeter said:
the distance once around the track is 500 meters = 1/2 kilometer

the track consists os two rectangular lengths = [MATH]2L[/MATH]and one circular length (two halves) = [MATH]\pi W[/MATH]
therefore the total length in meters is given by the equation in my previous post

you need to solve the equation for L in terms of W to get A = LW in terms of W

you also need to solve the equation for W in terms of L to get A = LW in terms of L
Click to expand...
I think I did that and I got the answer I put on top. Was your answer simplified? I don’t know what to do
 
[MATH]2L + \pi W = 500 \implies L = \dfrac{500-\pi W}{2}[/MATH]
[MATH]2L + \pi W = 500 \implies W = \dfrac{500-2L}{\pi}[/MATH]
 
What is confusing in this problem is that the width is NOT A RADIUS; it is a diameter.

So the perimeter of one semi-circle is

[MATH]\dfrac{1}{2} * 2 \pi * \dfrac{W}{2} = \dfrac{\pi W}{2}.[/MATH]
But there is a semi-cirle at each end. So the perimeter

[MATH]2L + 2 * \dfrac{\pi W}{2} = 2L + \pi W.[/MATH]
 
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