SOLUTION: A parkway 20 meters wide is spanned by a parabolic arc 30 meters long along the horizontal. If the parkway is centered, how high must the vertex of the arch be in order to give a m

Question 1012199: A parkway 20 meters wide is spanned by a parabolic arc 30 meters long along the horizontal. If the parkway is centered, how high must the vertex of the arch be in order to give a minimum clearance of 5 meters over the parkway?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A parkway 20 meters wide is spanned by a parabolic arc 30 meters long along the horizontal.
If the parkway is centered, how high must the vertex of the arch be in order to give a minimum clearance of 5 meters over the parkway?
:
that means the 30m arch starts and ends 5m on either side of the 20m highway
From the information given we can obtain the following x,y pairs
x=0; y=0
x=5; y=5
x=25; y=5
Using the form ax^ + bx + c = y; this parabola passes thru the origin, therefore c = 0
Two equations
x=5; y=5
25a + 5b = 5
:
x=25; y=5
625a + 25b = 5
simplify divide by 5
125a + 5b = 1
:
using elimination
125a + 5b = 1
25a + 5b = 5
---------------------subtraction eliminates b, find a
100a + 0 = -4
a = -4/100
a = -.04
:
Find b using eq 25a + b = 5
-.04(25) + 5b = 5
-1 + 5b = 5
5b = 5 + 1
b = 6/5
b = 1.2
:
the equation: y = -.04x^2 + 1.05x
Looks like this, green line is 5m height

:
"how high must the vertex of the arch be" find the axis of symmetry x = -b/(2a)
x =
x = 15
Find y when x=15
y = -.04(15^2) + 1.2(15)
y = -9 + 18
Vertex pair: 15, 9
y = 9 meters above the center of the high-way, blue line

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