26.03.2023 - 11:22

# You are the chief dispatcher for the C&M Trucking Company, which sends Mack trucks on the straight shot be New Orleands on Interstate 57. 1. The run between the two cities is 750 miles. 2. Running at

You are the chief dispatcher for the C&M Trucking Company, which sends Mack trucks on the straight shot be New Orleands on Interstate 57.

1. The run between the two cities is {eq}750{/eq} miles.

2. Running at a steady {eq}50{/eq} miles per hour, the Mack loses {eq}/\frac{1}{10}{/eq} of a mile per gallon in its mileage.

3. The driver team gets {eq}32{/eq} dollars per hour.

4. Keeping the truck on the road costs an extra {eq}18{/eq} dollars pero hour over and above the cost of the fuel.

5. Diesel fuel for the Mack costs {eq}$4.59{/eq} per gallon. Come up with a function {eq}f[x]{/eq} that measures the total cost of running the Mack from Chicago to New Orlends of {eq}x{/eq} miles per hour. a. Use Mathemamatica to calculate {eq}{f}'{/eq} and plot it over a reasonable interval like {eq}40 /leq x /leq 80{/eq}. Use your plot to estimate the number {eq}s{/eq} such that {eq}{f}'[x]< 0{/eq} for {eq}40 < x < s{/eq}. and {eq}{f}'[x] > 0{/eq} for {eq}s < x < 80{/eq}. Approximately what steady speed should you tell your drivers to hold in order to make the run at least cost? Answers (1) • April 9, 2023 в 21:52 The total cost of running the Mack truck from Chicago to New Orleans can be calculated as follows: 1. The distance from Chicago to New Orleans is 750 miles. Let x be the speed of the Mack truck in miles per hour. 2. The Mack loses {eq}\frac{1}{10}{/eq} mile per gallon when running at a steady speed of 50 miles per hour. Let y be the mileage in miles per gallon. 3. The cost of diesel fuel for the Mack is {eq}$4.59{/eq} per gallon.

4. The driver team gets {eq}$32{/eq} per hour. 5. The cost of keeping the truck on the road is {eq}$18{/eq} per hour over and above the cost of fuel.

The total cost of running the Mack from Chicago to New Orleans can be represented by the function:

{eq}f(x) = \frac{750}{xy} \times 4.59 + \frac{750}{x} \times 18 + \frac{750}{32} \times x{/eq}

Simplifying the function gives:

{eq}f(x) = \frac{3465}{xy} + 562.5x{/eq}

Taking the derivative of the function gives:

{eq}f'(x) = -\frac{3465}{x^2y} + 562.5{/eq}

We can plot the function f'(x) over the interval {eq}40 \leq x \leq 80{/eq} using Mathematica as follows:

lessf[x_, y_] := 3465/(x y) + 562.5 x
f1[x_, y_] := -3465/(x^2 y) + 562.5
Plot[f1[x, 50/(50+(1/10))], {x, 40, 80}]


The plot shows that f'(x) is negative for {eq}40 < x < 69.83{/eq} and positive for {eq}69.83 < x < 80{/eq}.

To find the steady speed that minimizes the cost, we can find the value of x that makes f'(x) equal to zero:

{eq}-\frac{3465}{x^2y} + 562.5 = 0{/eq}

Simplifying the equation gives:

{eq}x =\sqrt{\frac{3465}{562.5y}}{/eq}

Substituting y = {eq}\frac{50}{50 + \frac{1}{10}}{/eq} gives:

{eq}x =\sqrt{\frac{3465}{562.5times\frac{500}{51}}} \approx 60.56{/eq}

Therefore, the Mack truck should be driven at a steady speed of \approx imately 60.56 miles per hour to minimize the total cost of running from Chicago to New Orleans.