Designing a Bridge Arch (Math Problem)?

Question

Parabolic arches are known to have greater strength than other arches. A bridge with a supporting parabolic arch spans 60 ft. with a 30-ft. wide road passing underneath the bridge. In order to have a minimum clearance of 16 ft., what is the maximum clearance.

I don't know what minimum clearance is, so I assumed that it was the height from the ground to the bottom of the arch. Is this correct? If so, I still don't know how to solve the problem. So I still need your help. :)

Answer 1

An equation of y = kx^2 will not do for the parabolic arch of the bridge since this would have y = 0 in the centre and curve upward each side. The bridge would be sitting on the middle of the road!
You need an equation of the type y = a - bx^2 which curves the other way. You need to find the values of a and b.
You know that the bridge spans 60 ft which is 30 ft each side so when x = 30, y = 0 this gives you
0 = a - 900b (1)
You also know that when x = 15, y = 16 so that there is at least 16ft clearance across the whole road. This gives you
16 = a - 225b (2)
Putting (1) and (2) together as a pair of simulataneous equations gives you
a = 900b
16 = 900b - 225b = 675b ---> b = 16/675 = 0.0237
From this a = 900b = 900*0.0237 = 21.33
So the bridge equation is y = 21.33 - 0.0237x
Putting in x = 0 gives y = 21.33 and this is the height of the bridge at the centre or maximum clearance.

Answer 2

The peak of the arch will be above the center of the road. At the edge of the road, 15 ft from the center, the arch should be at least 16 ft above the road (that's what minimum clearance means in this problem). So find a value for k in the parabolic equation y=kx^2 such that y=16 when x=15. Then use that equation to find y when x=30.
Easier still, think: if y=kx^2=16 when x=15, what do you think happens to y when you double x?