Question

a doorway has the shape of a parabolic arch and is 9 feet high at the center and 6 feet wide at the base. if a rectangular box 5 feet high must fit through the doorway, what is the maximum width the box can have?

Explanation:

Step1: Set up the parabolic equation

Place the parabola with its vertex at the origin $(0,9)$. The general form of a parabola opening down - ward is $y = a(x - 0)^2+9=ax^{2}+9$. Since the parabola passes through the point $(3,0)$ (half of the 6 - foot base width and at ground - level), we substitute $x = 3$ and $y = 0$ into the equation: $0=a(3)^{2}+9$.

Step2: Solve for $a$

$0 = 9a+9$, then $9a=-9$, so $a=-1$. The equation of the parabola is $y=-x^{2}+9$.

Step3: Find the $x$ - value when $y = 5$

Set $y = 5$ in the equation $y=-x^{2}+9$. We get $5=-x^{2}+9$. Rearranging gives $x^{2}=9 - 5=4$. So $x = 2$ or $x=-2$.

Step4: Calculate the width

The width of the box is the distance between the two $x$ - values. The width $w=|2-(-2)| = 4$ feet.

Answer:

4 feet