Maths Challenge? - how many circle do i need to run in track and field for 1 mile
1) The running length of 440m will set includes a football field, whose shape is a rectangle with a semicircle at each end. If the area of the rectangle should be possible to find the lengths of its sides.
2) a window of a predetermined amount of cardboard will open with the radius c ^ 2, which is the maximum. Volume of the disk?
Thursday, January 28, 2010
How Many Circle Do I Need To Run In Track And Field For 1 Mile Maths Challenge?
3 comments:
Q1
440 = πx + 2y
Y = 220 - 0.5πx
A = xy
= 220x - 0.5πx ^ 2
A '= 220 - πx = 0
x = 220 / π = m 70.028175
Y = 110 m
Q2
This is a cut of the 4 corners of the bin? If yes explain:
We have a basic square, whose length and height 2p q. Our box have a size 2p x 2p x D.
c ^ 2 = (p + q ^ 2) + p ^ 2
q = √ (c ^ 2 - p ^ 2) - p> 0
Volume of the box, V (p)
= (2p) (2p) (q)
4p ^ 2 = [√ (c ^ 2 - p ^ 2) - p]
4p ^ 2 = √ (c ^ 2 - p ^ 2) - 4p ^ 3
V '(p) = 8p √ (c ^ 2 - p ^ 2) + (2p) (4p ^ 2) / 2 √ (c ^ 2 - p ^ 2) - 12p ^ 2
= [8p (c ^ 2 - p ^ 2) - 4p ^ 3] / √ (c ^ 2 - p ^ 2) - 12p ^ 2
= [8PC ^ 2 - 12p ^ 3] / √ (c ^ 2 - p ^ 2) - 12p ^ 2 = 0
√ (c ^ 2 - p ^ 2) = [8PC ^ 2 - 12p ^ 3] / 12p ^ 2
c ^ 2 - p ^ 2 = [(2c ^ 2 - 3p ^ 2) / 3p] ^ 2
9p ^ 2 (c ^ 2 - p ^ 2) = 4c ^ 4 to 12 (pc) ^ 2 ^ 4 + 9p
18p ^ 4 to 21 (pc) ^ 2 +4 c ^ 4 = 0
p ^ 2 = 21c ^ 2 / 36 + - √ [(21 c ^ 2) ^ 2 to 4 * 18 * 4 c ^ 4) / 36
= [+ 21 - √ 153] c ^ 2 / 36
= (0.583333 + - 0.343592) c ^ 2
p = + = + 0.9627697cop-0.4896337c
I remember that p is the length, sop> 0, q> 0
p = p = 0.9627697co 0.4896337c
but p = 0.9627697c give a negative value for q.
P should be as 0.4896337c.
V (p) p = 0.4896337c,
V (0.4896337c)
4p ^ 2 = √ (c ^ 2 - p ^ 2) - 4p ^ 3
0.3666069c = ^ 3
0.3666c unit cube = ^ 3
Q1
440 = πx + 2y
Y = 220 - 0.5πx
A = xy
= 220x - 0.5πx ^ 2
A '= 220 - πx = 0
x = 220 / π = m 70.028175
Y = 110 m
Q2
This is a cut of the 4 corners of the bin? If yes explain:
We have a basic square, whose length and height 2p q. Our box have a size 2p x 2p x D.
c ^ 2 = (p + q ^ 2) + p ^ 2
q = √ (c ^ 2 - p ^ 2) - p> 0
Volume of the box, V (p)
= (2p) (2p) (q)
4p ^ 2 = [√ (c ^ 2 - p ^ 2) - p]
4p ^ 2 = √ (c ^ 2 - p ^ 2) - 4p ^ 3
V '(p) = 8p √ (c ^ 2 - p ^ 2) + (2p) (4p ^ 2) / 2 √ (c ^ 2 - p ^ 2) - 12p ^ 2
= [8p (c ^ 2 - p ^ 2) - 4p ^ 3] / √ (c ^ 2 - p ^ 2) - 12p ^ 2
= [8PC ^ 2 - 12p ^ 3] / √ (c ^ 2 - p ^ 2) - 12p ^ 2 = 0
√ (c ^ 2 - p ^ 2) = [8PC ^ 2 - 12p ^ 3] / 12p ^ 2
c ^ 2 - p ^ 2 = [(2c ^ 2 - 3p ^ 2) / 3p] ^ 2
9p ^ 2 (c ^ 2 - p ^ 2) = 4c ^ 4 to 12 (pc) ^ 2 ^ 4 + 9p
18p ^ 4 to 21 (pc) ^ 2 +4 c ^ 4 = 0
p ^ 2 = 21c ^ 2 / 36 + - √ [(21 c ^ 2) ^ 2 to 4 * 18 * 4 c ^ 4) / 36
= [+ 21 - √ 153] c ^ 2 / 36
= (0.583333 + - 0.343592) c ^ 2
p = + = + 0.9627697cop-0.4896337c
I remember that p is the length, sop> 0, q> 0
p = p = 0.9627697co 0.4896337c
but p = 0.9627697c give a negative value for q.
P should be as 0.4896337c.
V (p) p = 0.4896337c,
V (0.4896337c)
4p ^ 2 = √ (c ^ 2 - p ^ 2) - 4p ^ 3
0.3666069c = ^ 3
0.3666c unit cube = ^ 3
Soccer field next to the square = 440m / (pi + 2) = 140m
If the second question requires more information than the box is rectangular or cylindrical
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