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?
First find the rule of the parabola:
We know that the parabola curves (is open) towards the bottom, so the rule is
y(x)=b-ax²
if we set the y-axis to cut the vertex.
Since
y(0)=16=b, so b=16
or y(x)=16-kx²
Also,
y(4)=0 tells us that k=1, or
the rule of the doorway is:
y(x)=16-x²
At
y(x)=8, we have
8=16-x²
Solve for x to get x=√8=2√2.
Thus the box will touch the doorway at
x=±2√2,
The maximum box size is double 2√2, or
4√2.
Well, it seems like we're in quite a "tight spot" with this question! So, we have a parabolic arch-shaped doorway with a height of 16 feet at the center and a base width of 8 feet. We need to find the maximum width of a rectangular box that is 12 feet high to fit through this doorway.
Now, imagine trying to fit our rectangular box through the arch. If the width of the box is greater than the width of the arch at any given height, it won't fit. So, we need to find the point on the arch where the width is equal to the width of the box.
Since the arch is a parabola, it has symmetry. That means if we draw a line vertically down the middle, the width of the arch will be the same distance from this line on both sides. So if we find the width at the center and subtract it from the total width of the arch, we'll have half of the maximum width of the box.
The width at the center of the arch is 8 feet, so half of the maximum width of the box is 8 feet. Therefore, the maximum width of the box can be 16 feet.
But hey, be warned, if you're planning to go through this archway with a "large and in charge" 16-feet wide box, you might just end up making a "grand entrance" and leaving a whole lot of chaos in your wake!
To find the maximum width the box can have to fit through the doorway, we need to determine the width of the doorway at a height of 12 feet.
Given that the doorway has the shape of a parabolic arch, we can consider it to be a symmetrical parabola. The standard equation of a parabola in vertex form is:
y = a(x - h)^2 + k
Where (h, k) represents the vertex of the parabola.
Since the doorway is 16 feet high at the center (vertex), we can write the equation for the parabolic arch as:
y = a(x - 0)^2 + 16
We also know that the doorway is 8 feet wide at the base. Therefore, at the base, the y-coordinate is 0. Substituting these values into the equation:
0 = a(8 - 0)^2 + 16
Expanding:
0 = 64a + 16
Rearranging the equation:
64a = -16
a = -16/64
a = -1/4
Now that we have the value of a, we can write the equation for the parabolic arch as:
y = (-1/4)x^2 + 16
To find the width of the doorway at a height of 12 feet (the height of the box), we need to find the x-coordinate. Substituting y = 12 into the equation:
12 = (-1/4)x^2 + 16
Rearranging the equation:
(-1/4)x^2 = -4
Multiplying both sides by -4:
x^2 = 16
Taking the square root of both sides:
x = ± 4
Since we are considering only the positive solution, the maximum width the box can have to fit through the doorway is 4 feet.
To find the maximum width the box can have, we need to determine the width of the doorway at a height of 12 feet.
Let's analyze the shape of the doorway. A parabolic arch is symmetric, meaning that the shape on one side of the arch is repeated on the other side. The highest point of the arch is called the vertex, which in this case is the center at a height of 16 feet. The base of the arch is 8 feet wide.
Since the shape is a parabola, we can use the equation of a parabola in vertex form: y = a(x - h)^2 + k, where (h, k) represents the vertex of the parabola.
In this case, the vertex is (0, 16), so the equation becomes: y = a(x - 0)^2 + 16, which simplifies to y = ax^2 + 16.
We know that the height of the doorway at the center is 16 feet (when y = 16 and x = 0). Plugging these values into the equation, we have: 16 = a(0)^2 + 16, which simplifies to 16 = 16a.
By solving this equation, we find that a = 1.
Now, we need to determine the width of the doorway at a height of 12 feet (y = 12). Plugging y = 12 into the equation y = ax^2 + 16, we get: 12 = 1x^2 + 16. Rearranging the equation, we have: x^2 = -4.
Since the width of the doorway cannot be negative, we can ignore the negative solution. Taking the square root of both sides, we get: x = √(-4).
However, since we are dealing with real-world measurements, √(-4) does not have a real solution. This means that there is no width for the box that will allow it to pass through the doorway when it is 12 feet high.
Therefore, the maximum width the box can have to fit through the doorway with a height of 12 feet is zero feet.