Home » Blog » Find the most economical speed for a truck driver subject to constraints

Find the most economical speed for a truck driver subject to constraints

A truck driver must drive 300 miles at a constant velocity of x miles per hour, where speed laws require 30 \leq x \leq 60. The truck consumes fuel at the rate of

    \[ 2 + \frac{x^2}{600} \]

gallons per hour, and the cost of the fuel is 0.3 dollars per gallon. The driver is paid D dollars per hour. Find the speed x at which the total cost of the trip is minimized when

  1. D = 0,
  2. D = 1,
  3. D = 2,
  4. D = 3,
  5. D = 4.

Before dealing with specific values of D, we write down an equation for the total cost of the trip in terms of the speed x. The driver must be paid D dollars per hour, and the trip takes 300/x hours (300 miles divided by the number of miles per hour), so we have

    \[ \text{Driver's wages } = \frac{300D}{x}. \]

Next, the total cost of the fuel is the number of hours of the trip (given by 300/x) times the number of gallons of fuel per hour times the cost of the fuel (30 cents per gallon):

    \[ \text{Fuel cost } = 0.3 \left(\frac{300}{x} \right) \left( 2 + \frac{x^2}{600} \right) = \frac{180}{x} + \frac{3x}{20}.\]

So the total cost of the trip in terms of the speed is

    \[ C(x) = \frac{300D + 180}{x} + \frac{3x}{20}. \]

Now for the specific values of D of the problem.

  1. If D = 0 we have

        \[ C(x) = \frac{180}{x} + \frac{3x}{20} \quad \implies \quad C'(x) = -\frac{180}{x^2} + \frac{3}{20}. \]

    Setting this derivative equal to 0 to find the critical points of C(x) we have

        \[ -\frac{180}{x^2} + \frac{3}{20} = 0 \quad \implies \quad x = 20 \sqrt{3} \text{ mph}. \]

    Since

        \begin{align*}  C'(x) &< 0 & \text{when} && x &< 20\sqrt{3} \\  C'(x) &> 0 & \text{when} && x &> 20 \sqrt{3}. \end{align*}

    Therefore, C(x) has a minimum at x = 20 \sqrt{3} mph. The cost at this speed

        \[ C(x) = \frac{180}{x} + \frac{3x}{20} = 6 \sqrt{3} = \$ 10.39. \]

  2. If D = 1 we have

        \[ C(x) = \frac{480}{x} + \frac{3x}{20} \quad \implies \quad C'(x) = -\frac{480}{x^2} + \frac{3}{20}. \]

    Setting this derivative equal to 0 to find the critical points of C(x), we have

        \[ -\frac{480}{x^2} + \frac{3}{20} = 0 \quad \implies \quad x = 40 \sqrt{2} \text{ mph}. \]

    Since

        \begin{align*}  C'(x) &< 0 & \text{when} && x &< 40 \sqrt{2} \\  C'(x) &>0 & \text{when} && x &> 40 \sqrt{2} \end{align*}

    the cost C(x) has a minimum at x = 40 \sqrt{2} mph. The cost at this speed is then

        \[ C = \frac{480}{x} + \frac{3x}{20} = 12 \sqrt{2} = \$ 16.97. \]

  3. If D = 2 we have

        \[ C = \frac{780}{x} + \frac{3x}{20} \quad \implies \quad C' = -\frac{780}{x^2} + \frac{3}{20}. \]

    Setting this derivative equal to 0,

        \[ -\frac{780}{x^2} + \frac{3}{20} = 0 \quad \implies \quad x = 20 \sqrt{13}. \]

    However, in this case 20 \sqrt{3} > 60, and the problem tells us that speed laws require 30 \leq x \leq 60. Since the cost is decreasing on the interval 30 \leq x \leq 60 (since the derivative is negative everywhere on the interval) we take x = 60 mph. Then,

        \[ C = 13 + 9 = \$ 22. \]

  4. If D = 3, then just as above we take x = 60 mph and obtain

        \[ C = \frac{1080}{60} + 9 = \$ 27. \]

  5. If D = 4, we again have x = 60 mph and get the minimal cost as

        \[ D = \frac{1380}{60} + 9 = \$ 32. \]

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):