algebra help?

Question

A 1/4 -mile track for racing stock cars consists
of two semicircles connected by parallel straightaways (see
the figure). In order to provide sufficient room for pit crews,
emergency vehicles, and spectator parking, the track must enclose an area of 100,000 square feet. Find the length of the
straightaways and the diameter of the semicircles to the nearest foot. Recall: The area A and circumference C of a circle of diameter d are given by A= pie times diameter squared divided by 4 and c= pie times diameter.


i do not know how to get it into an equation although i have been shown some suggestions and even once into the equation i do no know how to solve... can someone please fully help me...

i do not know how to work problems with the substitution method, if someone would be so kind to work it out maybe i can see and learn... I am a visual learner

here is the diagram
http://i7.photobucket.com/albums/y294/joetta/cart.jpg

can someone show me the steps to solve it from the point when u have two equations.. i am just as confused with the two equations as i am the problem

Answer 1

A = 100000 = пD²/4 + DL
C = 5280/4 = 1340 = пD + 2L

There are your two equations. Now solve.

пD + 2L = 1340
L = (1340 - пD)/2

пD²/4 + DL = 100000
L = (100000 - пD²/4)/D

(100000 - пD²/4)/D = (1340 - пD)/2
200000 - пD²/2 = 1340D - пD²
пD² - пD²/2 = 1340D - 200000
пD²/2 = 1340D - 200000
D² - 2680D/п + 400000/п = 0
D² - 853D + 127323 = 0

D = 192, 660

Since L = (1340 - пD)/2
L = 368, -366

Thus the only solution is D = 192 and L = 368

Hope I calculated this right.

Answer 2

The area inside the track is one circle of diameter X.

The rectangle is X*Y

For the area you are given

100,000 ft**2= X*Y+pi*X/4

For the circumference you are given

1/4mi = 5280ft/4 = 2*X + 2*Y + pi*X

You now have two equations that you solve simutaneously.

If you are unfamilar with substitution methods let's use the cicumference equation and solve Y in terms of X.

1/4mi = 5280ft/4 = 2*X + 2*Y + pi*X
or 1320 ft = 2Y + X(2+pi)
then
2Y = 1320 ft - X(2+pi)
or
Y = (660 -X(2+pi)) ft

we can now substitute this expression into the area equation every Y appears and we can solve for the single variable X.

so

100,000 ft**2= X*Y+pi*X/4

becomes

100,000 ft**2 = X*[(660 -X(2+pi)) ft] + X*pi/4 or

100,000 ft**2 = X*[(660 -X(2+pi)) + pi/4]ft or

0 = [pi-2]*X**2 + 660*X + [pi/4-100,000]

which you can solve with the quadratic equation. But I'll skip that extra step and show you another way that elimnated the need for the quadratic equation.

The other method (often easier) is simultaneous solutions. If you can see how to eliminate one variable from both equations (by multiplying the equation by scalers and adding/subtracting the two equation together to get one variable eliminated alltogether) you then have a 'combined' equation with a single variable just as you did with the substitution method.

For these equations
100,000 ft**2= X*Y+pi*X/4 and 1320 ft = 2*X + 2*Y + pi*X

you first divide the area equation by X (this also helps with unity... gets both equation in feet) and the second equation by 2.

(100000ft**2)**1/2X = Y + pi/4 for the area/x
660 ft = Y +X(pi/2 + 1) for the circ/2

subtract the top from the bottom

(660-100,000**1/2)ft = X(pi/2 + 1) - pi/4

solve for x

X= [(660-100,000**2)+pi/4]/[(pi/2 + 1)]

either substitution or simultaneous solving gives you the same answer. It sounds like you needed to show the substitution method though so go that route for homework's sake.

If substitution was not mentioned, then practice simultaneous methods. When you get to 3 (or more) variables the substitution method gets a bit tedious, and the simultaneous method really starts to save you some laborious work.