Question 1035950: The maximum height of the arch is 21feet and the width of the roadway, including shoulder lanes, is 42 feet. At the edge of the left shoulder lane, the height of the arch is 15.4 feet. Determine the width of the arch to the nearest tenth of a foot.
Found 2 solutions by ankor@dixie-net.com, josmiceli:
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! The maximum height of the arch is 21feet and the width of the roadway, including shoulder lanes, is 42 feet.
At the edge of the left shoulder lane, the height of the arch is 15.4 feet. Determine the width of the arch to the nearest tenth of a foot.
:
The arch is in the shape of a parabola. Create an equation that has these requirements.
ax^2 + bc + c = y
We will let the axis of symmetry go thru the origin then c = 21 ft
Write two equations
x = -21; y = 15.4
(-21^2)a - 21b + 21 = 15.4
441a - 21b = 15.4 - 21
441a -21b = -5.6
and
x = +21; y = 15.4
441a + 21b + 21 = 15.4
441a + 21b = -5.6
:
Use elimination, add these two equations
441a - 21b = -5.6
441a + 21b = -5.6
----------------------Addition, eliminates b find a
882a = -11.2
a = -11.2/882
a = -.0127
Find b
441(-.0127) + 21b = -5.6
-5.6 + 21b = -5.6
21b = 0
b = 0
Here is our equation
y = -.0127x^2 + 21
Looks like this
Green line = 15.4, note the width at this point is about 42 ft (21+21)
:
"Determine the width of the arch to the nearest tenth of a foot."
solve for y = 0
-.0127x^2 + 21 = 0
-.0127x^2 = -21
x^2 = -21/-.0127
x^2 = +1653.5
x = +/-
x = +40.66
and
x = -40.66
Find the total width:
40.66 - (-40.66) = 81.3 ft is the total width
Answer by josmiceli(19441)  (Show Source):
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