SOLUTION: A rectangular object 25 m wide is to pass under a parabolic arch that has a width of 32 m at the base and a height of 24 m at the center. If the vertex of the parabola is at the to

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: A rectangular object 25 m wide is to pass under a parabolic arch that has a width of 32 m at the base and a height of 24 m at the center. If the vertex of the parabola is at the to      Log On


   

Question 1196473: A rectangular object 25 m wide is to pass under a parabolic arch that has a width of 32 m at the base and a height of 24 m at the center. If the vertex of the parabola is at the top of the arch, what maximum height should the rectangular object have?
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The vertex is to be at the top of the arch; let the origin of a coordinate system be at the center of the base of the arch.

With a height of 24 at the center of the arch, the coordinates of the vertex are (0,24).

And with a width of 32 at the base, the coordinates of the endpoints of the arch are (-16,0) and (16,0).

With those conditions, the equation of the parabolic arch is y=ax%5E2%2B24, where a is a (negative) constant to be determined.

That constant can be determined using either endpoint of the arch.

0=a%2816%5E2%29%2B24
0=256a%2B24
a=-24%2F256=-3%2F32

The equation of the arch is y=%28-3%2F32%29x%5E2%2B24

The object that is to pass through the arch has a width of 25, so it will extend 12.5 units each side of the center of the arch. To find the maximum height of the object, evaluate the equation for x=12.5. The answer is an ugly number, so I'll let you do that part. You might want to use a graphing calculator instead of working with the ugly arithmetic....