SOLUTION: a parabolic opening of a tunnel is 32m wide measured from side to side along the ground.At the points that are 4m from each side, the tunnel entrence is 6m high find the maximum

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: a parabolic opening of a tunnel is 32m wide measured from side to side along the ground.At the points that are 4m from each side, the tunnel entrence is 6m high find the maximum      Log On


   

Question 1100845: a parabolic opening of a tunnel is 32m wide measured from side to side along the ground.At the points that are 4m from each side, the tunnel entrence is 6m high
find the maximum height to the tunnel to 1 decimal place
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
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Put the origin of the coordinate system at the highest point of the arch, the axis "y" directed vertically up.

so the the origin of the coordinate system will be THE VERTEX of the parabola.


You have y = ax^2 as the equation of the parabola with the negative unknown coefficient "a".


The equation for "a" is

y(12) - y(16) = a%2A12%5E2 - a%2A16%5E2 = 6,     (1)

saying that  at the distance of 12 = 16-4 meters from the side the height of the tunnel is 6 meters.


Then  a = 6%2F%2812%5E2-16%5E2%29 = 6%2F%28144-256%29 = -6%2F112 = -3%2F56.


Hence and finally, the equation of the parabola is

y = %28-3%2F56%29%2Ax%5E2 

in this coordinate system.   


The maximal height of the tunnel is  %283%2F56%29%2A16%5E2 = 13.7 meters.



Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


I prefer a different setup than the other tutor for solving this kind of problem. Look at the two solution methods and use the one which "works" better for you.

I put the origin of the coordinate system at the middle of the tunnel, at "ground" level. Since the tunnel is 32m wide, the x-intercepts are at (-16,0) and (16,0).

And then the equation for the parabola is y+=+a%28x-16%29%28x%2B16%29

The other known points on the parabola are at a height of 6m, 4m in from the edges of the tunnel -- at (12,6) and (-12,6).

Then find the value of the constant a by plugging the coordinates of one of those points into the equation:
6+=+a%2828%29%28-4%29
6+=+-112a
a+=+-3%2F56

The equation of the parabola is y+=+%28-3%2F56%29%28x-16%29%28x%2B16%29

The maximum value is at x=0: %28-3%2F56%29%28-256%29+=+96%2F7