Question

The front of an aircraft hanger is being built in the shape of a parabola, which is 32 ft. wide, and has a maximum height of 16 ft., AND must have a rectangular doorway that is 8 ft. tall. What is the maximum width of the doorway?

Answer 1

Answer:

22.6 ft

Step-by-step explanation:

We are given that

Width of parabola=32 ft

Half width of parabola=$\frac{32}{2}=16ft$

Distance from origin on right side on x-axis=16ft

Distance from origin on left side =-16 ft

Maximum height of parabola=16 ft

Therefore, the point (0,16) lie on the parabola.

Equation of parabola along y-axis is given  by

$y=a(x-h)^2+k$

Where vertex=(h,k)

Vertex of parabola=(0,16)

Substitute the value of vertex

$y=ax^2+16$..(1)

Equation(1) is passing through the point (16,0)

Therefore, $0=a(16)^2+16$

$-16=256a$

$a=-\frac{16}{256}=-\frac{1}{16}$

Substitute the value of a in equation(1)

$y=-\frac{1}{16}x^2+16$

Height of doorway=8 ft

It means we have to find the value of x at y=8

Substitute the value of y

$8=-\frac{1}{16}x^2+16$

$8-16=-\frac{1}{16}x^2$

$-8=-\frac{1}{16}x^2$

$x^2=8\times 16=128$

$x=\sqrt{128}=11.3$ft

Width of rectangular doorway=2x=2(11.3)=22.6 ft

Hence, the width of rectangular doorway=22.6 ft