I\'ve been stuck on this problem and I can\'t seem to understand it! I would gre

Question

I've been stuck on this problem and I can't seem to understand it! I would greatly appreciate any help! Thanks!
William, a logistician needs 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 notice prior to use. If the tunnel is roughly modeled by f(x)=-0.1x^2+3.2x-3.5, should William make arrangements for the train to use the tunnel? Show work that supports your conclusion.    I've been stuck on this problem and I can't seem to understand it! I would greatly appreciate any help! Thanks!
William, a logistician needs 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 notice prior to use. If the tunnel is roughly modeled by f(x)=-0.1x^2+3.2x-3.5, should William make arrangements for the train to use the tunnel? Show work that supports your conclusion.   
William, a logistician needs 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 notice prior to use. If the tunnel is roughly modeled by f(x)=-0.1x^2+3.2x-3.5, should William make arrangements for the train to use the tunnel? Show work that supports your conclusion.   

Explanation / Answer

The graph of the equation is a parabola.
Because the second degree term has a negative coefficient, the parabola opens downward.
The x component of the vertex = -b/(2a)

                                            = -3.2/(2*(-0.1))

                                            = -3.2/(-0.2)

                                            = 16
The y component of the vertex = 0.1(16)^2 + 3.2(16) - 3.5      ( put x=16 in equation of parabola)
                                            = 25.6 + 51.2 - 3.5                                              

                                           = 22.1
With the widest point being 10 feet, that means 5 feet either side of x = 16.
One side is x = 16 - 5 = 11 feet and the other side is x = 16 + 5 = 21 feet.

Now we need the height of the opening at the ends of this 10 foot width.

So for x = 11
y = -0.1x^2 + 3.2x - 3.5
y = -0.1(11)^2 + 3.2(11) - 3.5
y = -0.1(121) + 35.2 - 3.5
y = -12.1 + 35.2 - 3.5
y = 19.6 feet high when x is at the 11 foot width mark

And for x = 21
y = -0.1x^2 + 3.2x - 3.5
y = -0.1(21)^2 + 3.2(21) - 3.5
y = -0.1(441) + 67.2 - 3.5
y = -44.1 + 67.2 - 3.5
y = 19.6 feet high when x is at the 21 foot width mark

So the difference between x = 11 and x = 21 is 10 feet. But at the ends of the 10 foot width, the tunnel height is 19.6 feet. Assuming that the train's height is the same for the entire 10 foot width, then the ends are 0.4 feet too short for a 20 foot tall train to fit.

Therefore, the train needs a different route.

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