William, a logistician need to route a freight train that is 20 feet at its tallest point and 10 feet at its widest point within 3 days. the most direct path includes a single-track tunnel that needs 24 hour notic prior to use. if the tunnel is roughly modeled by

Should William make arrangements for the train to use the tunnel?

-3.2/2(-0.1)=16

F(16)=-0.1(16)^2 +3.2(16)-3.5=22.1

22.1>20

Now I find the width:

16+5=21 16-5=11

f(21)=-0.1(21)^2+3.2(21)-3.5=19.6

f(11)=-0.1(11)^2+3.2(11)-3.5=19.6

Because 19.6<20 the train cannot fit. Is this correct?

Thank you in advance.

### 1 Answer | Add Yours

The tunnel is modeled by a concave down parabola defined as:

`f(x)=-0.1x^2+3.2x-3.5`

If we assume that the tallest point of the train is at the centre then first we want to find what the maximum value of f(x) is. To do so, first take the derivative, set it to 0, and solve for x:

`f'(x)=-0.2x+3.2=0 -gt 0.2x=3.2 -gt x=16`

Now, find the value for f(x) at the maximum point x=16:

`f(16)=-0.1(16)^2+3.2(16)-3.5=22.1`

Therefore, the tunnel is tall enough to accomodate the train because 22.1>20.

If we assume that the widest point of the train is at its center, than we wish to find how wide the parabola is at y=20/2=10

`10=-0.1x^2+3.2x-3.5`

`0=-0.1x^2+3.2x-3.5-10`

`0=-0.1x^2+3.2x-13.5`

We can use the quadratic forumla to solve for x:

`a=-0.1; b=3.2; c=-13.5`

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`x=(-3.2+-sqrt(3.2^2-4(-0.1)(-13.5)))/(2(-0.1))`

`x_1=(-3.2+sqrt(4.84))/-0.2=5`

`x_2=(-3.2-sqrt(4.84))/-0.2=27`

Finally, we want to know what the distance is between `x_1` and `x_2` :

`d=x_2-x_1=27-5=22`

Therefore, the tunnel is wide enough to accomodate the maximum train width because 22>10.

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