Question 1206768
<pre>
Let the parabola have vertex (0,27), and let the 50-foot parkway extend
from (-25,0) to (25,0). To have minimum clearance at the edges of 15 feet,
the parabola must pass through points (-25,15) and (25,15).

{{{drawing(800,470, -40,40,-10,37,

graph(800,470, -40,40,-10,37,

(-(12/625)x^2+27)*(sqrt(x+37.5)/sqrt(x+37.5))*(sqrt(37.5-x)/sqrt(37.5-x))),
line(-25,0,-25,15), line(25,0,25,15), line(-25,0,25,0),
locate(26,16,"(25,15)"), locate(-31.5,16,"(-25,15)"),
locate(1,29,"(0,27)"),locate(26,2,"(25,0)"), locate(-31.5,2,"(-25,0)") 
)}}}

The equation of a parabola has several forms,  I'll use this one:

{{{(x-h)^2}}}{{{""=""}}}{{{4a(y-k)}}}

where (h,k) is the vertex (0,27)

{{{(x-0)^2}}}{{{""=""}}}{{{4a(y-27)}}}

{{{x^2}}}{{{""=""}}}{{{4a(y-27)}}}

Since it passes through (25,15), we substitute

{{{25^2}}}{{{""=""}}}{{{4a(15-27)}}}
{{{625}}}{{{""=""}}}{{{4a(-12)}}}
{{{625}}}{{{""=""}}}{{{-48a}}}
{{{625/(-48)}}}{{{""=""}}}{{{a}}}
{{{-625/(48)}}}{{{""=""}}}{{{a}}}

So the equation of the parabola is

{{{x^2}}}{{{""=""}}}{{{4(-625/48)(y-27)}}}

{{{x^2}}}{{{""=""}}}{{{expr(-625/12)(y-27)}}}

To find the width of the entire arch at the
bottom, we must find the x-intercepts. So we 
substitute 0 for y:

{{{x^2}}}{{{""=""}}}{{{expr(-625/12)(0-27)}}}
{{{x^2}}}{{{""=""}}}{{{16875/12}}}
{{{x}}}{{{""=""}}}{{{"" +- sqrt(1406.25)}}}
{{{x}}}{{{""=""}}}{{{"" +- 37.5}}}

So the width of the arch at the bottom goes from
(-37.5,0) to {37.5,0). 

And so the width of the arch at the bottom is 

(37.5)(2) = 75 feet.

Edwin</pre>