Question

Help Me! A doorway has the shape of a parabolic arch and is 16 feet high at the center and 8 feet wide at the base. If a rectangular box 12 feet high must fit through the doorway, what is the maximum width the box can have?

Answer 1

What grade level is this? This isn't calculus, is it? (Just making sure.)

Answer 2

What's your equation for the doorway?

Answer 3

it is pre cal

Answer 4

y=ax^2+16 i think

Answer 5

"12 feet high" Ah, missed that bit. This takes it out of the realm of calculus (or Grade 11 Honors... our teacher loves to troll us).

--

OK, so for your equation, you've solved it part of the way.

We know that the parabola touches the ground whenever \( y=0 \).

Since the width is 8, we know it will touch the ground at (0,0) and (2 * 8,0).

In other words, if x = 0 or x = 16, then y = 0.

\(\large y = \ \ \ 0 = a (0)^2 +16 \)
\(\large y = \ \ \ 0 = a (16)^2 +16 \)

Solve for \( a \).

Answer 6

a=-1/16?

Answer 7

Yep.

So we have our equation of the parabola:

\[\Large y = \frac{-1}{16}x^2 +16\]

Recall that the height is 12. That means that \(y=12\). With me so far?

Answer 8

yea

Answer 9

Sorry, I just realized we made a mistake. The equation for the doorway should pass through the points (0,0) and (4,16) and (8,0).

To find the equation of the doorway, we know that the roots of the equation are 0 and 8 so:
\[\large y = a(x-0)(x-8)\]
We then find \(a\) using (4,16):
\[\large {y \ \ = a(x-0)(x-8)} \]\[\large {16 = a(4-0)(4-8) \\ 16 = a(4)(-4) \\ 16 = -16a \\ \ \\ a = -1} \]

So we have our equation for the doorway:
\[\large y = -1(x-0)(x-8) \\ \large \ \ = -x(x-8)\]

Phew.

Answer 10

Since we know the height of the box is 12, we know it collides with the doorway at \(y=12\).



\[\large {y \ \ \ = -x(x-8) \\ 12 = -x(x-8)}\]

Solve for \( x \).