Precal Word Problem about Parabola and Doorway Width

Question

The front of an aircraft hanger is being built in the shape of a parabola, which is 162 ft. wide, and has a maximum height of 95 ft. AND must have a rectangular doorway that is 60 ft. tall. What is the maximum width of the doorway? (Round your answer to the nearest tenth of a foot.)

Answer 1

Graph the parabola with vertex at (0,95). x-intercepts at (-81,0,) and (81,0)

find the equation of that graph.

y-h = -a(x-k)^2 where (h,k) = (0,95) the vertex

y-95=-a(x)^2

plug in either x-intercept and solve for a8

0-95 =-a(81)^2

a = 95/81^2 = 95/6561) =

y-95 = -(95/6561)x^2

substitute 60 for you, and solve for x, half the maximum height of the door

60-95 = -(95/6561)x^2

x^2 = 35 x 6561/95

x = 81sqr35/95 = 81sqr7/19= 81x0.607=49.17 feet for half the door width

multiply by 2

98.34 feet maximum width of door

just a rough check, substitute the point (50,60) into the equation for the parabola

60-95 = -(95/81^2)(50)^2

-35 =-95x2500/81^2 = -35 rounded to an integer.

we should have used 49..17 instead of 50, but that made the calculations cumbersome.

so using 50 we got slightly more than 35 or slightly less than -35