A highschool is planning to build a new playing field surrounded by a running track. The track coach wants two laps around the track to be 1000m. The football coach wants the rectangular infield area to be as large as possible. Can both coaches be satisfied?
can somebody plz help me this is really important and any help will be greatly appricieted.
thnx a lot.
grade 11:
2pir + 2w = 500m
w = 250 - pir
Area of in = x(2r)
= (250 - pir)2r
it's a "partially factored" quadratic...
0 = (250 - pir)2r
r = 0, 250/pi
(0 + (250/pi))/2 = 125/pi
r = 125/pi
...
w = 125
250/pi m by 125 m
let the length of the rectangular infield be x m
let the width of the rectangular infield be 2r m ,(making the radius of the semicircular ends r m)
once around is 500 m
2x + 2pir = 500
x + pir = 250 or x = 250 - pir
Area of infield = x(2r)
= 2r(250 - pir)
= 500r - 2pir^2
dA/dr = 500 - 4pir = 0 for a max of A
r = 39.788
x = 250 - pi(39.788)
= 125
so the length of the infield should be 125 m
and its width should be 79.58 m
(Wow, what a field)
it 125/ pi and 125 m :D
the answer to this question can not use derivatives (or shouldn't) because it is in a grade 11 math textbook within a quadratics unit. Anyone else want to give it a go?
Well, well, well, it seems like we have a little conflict between the track coach and the football coach. Let's see if we can bring some clownish humor to settle this debate.
Now, imagine the two coaches sitting at a table, trying to find a solution. The track coach says, "I want two laps around the track to be 1000 meters!" And the football coach responds, "But I want a big rectangular infield area for our football games!"
I'd say the best way to satisfy both coaches is to think outside the box, or rather, outside the track. How about a compromise? Instead of building a traditional rectangular field inside the track, we can create a D-shaped playing area. This way, the track coach gets his desired two laps equaling 1000 meters, while the football coach gets a larger infield area.
And imagine the possibilities! The football team could set up some unique plays, navigating around the curved edges of the D-shaped field. Plus, it could make for some hilarious moments watching players chase the ball around those bends. Talk about a game with a twist!
So, my dear friend, I believe there's a solution where both coaches can be satisfied. Just remember, a little creativity can go a long way when solving problems. Good luck, and may your laughs be plentiful on this twisted field of compromise!
To solve this problem, we need to consider the dimensions of the playing field and its surrounding track. Let's assume the rectangular infield has lengths "l" (along the longer sides) and "w" (along the shorter sides).
The track coach wants two laps around the track to be 1000m. This means the perimeter of the track (2 laps) should be 1000m. Since a rectangular infield has two lengths and two widths, the perimeter of the track can be expressed as:
Perimeter of track = 2l + 2w
The football coach wants the rectangular infield area to be as large as possible, which means maximizing the product of the length and width:
Infield area = l * w
Now, let's analyze the requirements:
1. Two laps around the track should be 1000m:
2l + 2w = 1000
2. The infield area should be as large as possible:
Infield area = l * w
We can solve this problem using a method called optimization. To start, let's solve equation 1 for l:
2l + 2w = 1000
2l = 1000 - 2w
l = (1000 - 2w) / 2
l = 500 - w
Now substitute this expression for l into the equation for the infield area:
Infield area = (500 - w) * w
Infield area = 500w - w^2
To find the maximum area, we can take the derivative of the area with respect to w, set it equal to 0, and solve for w:
d(Infield area) / dw = 500 - 2w = 0
500 = 2w
w = 250
Substituting this value of w back into the equation for the infield area:
Infield area = (500 - 250) * 250
Infield area = 250 * 250
Infield area = 62500 m^2
Therefore, the maximum infield area that satisfies both coaches is 62500 square meters.
In conclusion, based on the given constraints, it is possible for both coaches to be satisfied by constructing a playing field with an infield area of 62500 square meters and a track perimeter of 1000 meters.