# Should william make arrangements for the train to use the tunnel in the following case?William, a logistician need to route a freight train that is 20 feet at its tallest point and 10 feet at its...

Should william make arrangements for the train to use the tunnel in the following case?

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

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### 3 Answers

The height of the tunnel that is being considered for use has a height given by the function `f(x) = -1*x^2 + 3.2*x - 3.5`

The minimum value of f(x) is at the solution of `f'(x) = 0` and if the solution is x = a, `f''(a) < 0` .

For f(x) = -1*x^2 + 3.2*x - 3.5, f'(x) = -0.2*x + 3.2 and f''(x) = -0.2 which is negative for all values of x.

Solving f'(x) = 0 gives -0.2*x + 3.2 = 0

=> x = 3.2/0.2 = 16

f(16) = 22.1

The minimum height of the tunnel is 22.1 ft. The maximum height of the train which is 20 ft is lesser than 22.1 ft.

**As a result the track can be used by the train.**

### User Comments

The height of the tunnel that is being considered for use has a height given by the function

The minimum value of f(x) is at the solution of and if the solution is x = a, .

For f(x) = -1*x^2 + 3.2*x - 3.5, f'(x) = -0.2*x + 3.2 and f''(x) = -0.2 which is negative for all values of x.

Solving f'(x) = 0 gives -0.2*x + 3.2 = 0

=> x = 3.2/0.2 = 16

f(16) = 22.1

The minimum height of the tunnel is 22.1 ft. The maximum height of the train which is 20 ft is lesser than 22.1 ft.

As a result the track can be used by the train.what is the width? or how do you solve the width?

To find the width of the tunel you need to find the x-intercepts.

The height of the tunnel that is being considered for use has a height given by the function

The minimum value of f(x) is at the solution of and if the solution is x = a, .

For f(x) = -1*x^2 + 3.2*x - 3.5, f'(x) = -0.2*x + 3.2 and f''(x) = -0.2 which is negative for all values of x.

Solving f'(x) = 0 gives -0.2*x + 3.2 = 0

=> x = 3.2/0.2 = 16

f(16) = 22.1

The minimum height of the tunnel is 22.1 ft. The maximum height of the train which is 20 ft is lesser than 22.1 ft.

As a result the track can be used by the train.

what is the width? or how do you solve the width?