- Download PDF
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?
I think the height is 22.1 ft. and the width is about 29.74ft
So, the train should fit. Is this correct?
2 Answers | Add Yours
29.74 is the width of the tunnel at the very bottom, so you are assuming that the maximum width of the train is at the bottom of the train. As I said, you have to make all kinds of assumptions to solve this problem, you can't say either way whether the train truely fits.
In truth, there isn't enough information provided by the question to do any exact calculations on whether the train will fit. In order to do so we need to know the exact shape of the train. Just because the maximum height and width may fit, that doesn't mean that the rest of the train will depending on how the rest of it is shaped. Instead, when I answered your previous question I assumed for simplicity that the maximum height and width was at the center of the train.
With this information we found that the maximum height was at x=16 and that f(16) of the function defining the tunnel was 22.1, which is greater than 20, and therefore the train's maximum height should fit.
Next we found that at y=10 (half of the maximum height), that the values for x that equate to f(x)=10 were 5 and 27. The distance between these two points is 22, which is larger than 10. Therefore, the maximum width should fit.
I just have one more question regarding this problem.
When I graph the function f(x)=-0.1x^2 +3.2x-3.5, The height is 22.1ft. and the x-intercepts are about 30.87 and 1.13, so the width of the tunnel would be 29.74, correct? When I graph the equation 0=-0.1x^2+3.2x-13.5, The height becomes 12.1 and the width becomes 22ft. So, 29.74ft. is the width of the tunnel, right? Because 29.74 ft.>10 the train should fit.
Is this also correct? If not, please explain.
Thank you for your patience.
We’ve answered 320,195 questions. We can answer yours, too.Ask a question