Expert: Sherry Wallin - 10/25/2010

Question
Hi,
I need to do some homework quickly - Can you please help?
It starts like this ...
2 arches (like MacDonalds)which are identical, both 8 metres tall and 6 metres wide, each allowin only one lane of traffic through. All vehicles will enter the property through the left arch and exit through the right arch.
There are different vehicles of different sizes...
1. What is the maximum width (to the nearest centimentre) of a truck that is 4.8 metres tall if it is to fit through the arches?
2. What is the maximum height (to the nearest centimetre ofa truckthat is 4.5 metres if it is to fit through the arches?
3. What is the maximum height and width (to the nearest centimetre( of a truck if it is to fit through the arches, given that the truck's height and width is the same?

Answer
Dianne~

The first thing you will need to do is write an equation that describes the arches. The arches are a parabola which means you will need to write the equation of a parabola which is a 2nd degree equation, i.e., a quadratic equation. You are given enough information to determine this equation. First you are told that the height is 8m and that the width is 6m. Place the arch at the origin so that the parabola starts at (0,0) and ends at (6.0). That means the midpoint (6-0)/2 = 3 is the x coordinate of the vertex while the y coordinate is 8. So there you have 3 sets of points and you can solve a system to find the values of each of the coefficients of the quadratic equation:

The standard 2nd degree (quadratic) equation looks like:
y = ax^2 + bx + c

First use the point (0,0) to find c:
0 = a*0^2 + b*0 + c = 0 + 0 + c => c = 0
So now you know your quadratic equation will look like
y = ax^2 + bx

Now use the point (6,0) to get an equation in a and b:
0 = a*6^2 + b*6 => 36a + 6b = 0  call this equation 1

Use the vertex for another quadratic equation:
8 = a*3^2 + b*3 => 9a + 3b = 8   call this equation 2

multiply equation 2 by -2 and add to equation 1:
-18a -6b = -16
36a + 6b = 0   add these two
18a = - 16 => a = -8/9

substitute a = -8/9 into equation 1:
36(-8/9) + 6b = 0 => -32 + 6b = 0 => 6b = 32 => b = 16/3

You now know the equation that describes the arches:
y = (-8/9)x^2 + (16/3)x

To answer #1, think of drawing a horizontal line at a height of 4.8m,
the x coordinate is still 3 there so the point is (3,4.8). You want to know what the width is, or the length of that line. Everywhere on that line y is 4.8 so substitute y = 4.8 into the equation:

4.8 = (-8/9)x^2 + (16/3)x => (-8/9)x^2 + (16/3)x - 4.8 = 0

using whatever method you want find out what the values of x are and there will be two of them and they will be the edges of the parabola where the line y = 4.8 crosses the parabola. I used the quadratic formula and found the x coordinates to be 1.1025 and the other 4.8975
and we want the length of that line from 1.1025 to 4.8975 which is
3.795m which is the maximum width of a vehicle to pass through the arches. Notice they want centimeters so multiply 3.795 by 100 to get 379.5 cm which the nearest cm is 379cm. Normally you would round up but we want the vehicle to fit so we will round down.

For the 2nd question I am assuming they are saying now that the width is 4.5m. that means let x = 4.5 and solve for y, the height:
y = (-8/9)(4.5)^2 +(16/3)(4.5) => y = -18 + 24 = 6 => y = 6
The maximum height of a vehicle having a width of 4.5m is 6m which is 600 cm.

For the 3rd question: You want to know the solution to the equation:
y = (-8/9)x^2 + (16/3)x so that x = y so solve that equation:
x =  (-8/9)x^2 + (16/3)x and you will find that x = y = 4.875
This means that a truck that is 487cm wide by 487cm tall will fit through the arches.

Math Prof