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

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

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.

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

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

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

$C = 13 + 9 = \$ 22.$
If $D = 3$ , then just as above we take $x = 60$ mph and obtain
$C = \frac{1080}{60} + 9 = \$ 27.$
If $D = 4$ , we again have $x = 60$ mph and get the minimal cost as
$D = \frac{1380}{60} + 9 = \$ 32.$