d. let l = 400 mi, p = $4/gal, and w = $20/hr. at what (constant) speed should the vehicle be driven to…

Explanation:

Step1: Establish cost - speed relationship

Let the speed of the vehicle be $v$ (in mph). The time taken for the trip $t=\frac{L}{v}=\frac{400}{v}$ hours. Assume the fuel - consumption rate is a function of speed, say $f(v)$ (gal/hour). A common model for fuel - consumption rate is $f(v)=av^{2}+bv + c$. For simplicity, if we assume a linear relationship between fuel consumption and speed in a certain range and use the fact that the total cost $C$ of the trip is the sum of fuel cost and labor cost. The fuel cost is $p\times f(v)\times t$ and the labor cost is $w\times t$. Let's assume a simple model where the fuel consumption rate $f(v)=\frac{k}{v}$ (a more realistic model could be more complex, but this is a basic one for illustration). The total cost $C(v)=p\times f(v)\times t+w\times t$. Substituting $L = 400$, $p = 4$, $w = 20$ and $t=\frac{400}{v}$, we get $C(v)=4\times\frac{k}{v}\times\frac{400}{v}+20\times\frac{400}{v}=\frac{1600k}{v^{2}}+\frac{8000}{v}$. If we assume that the fuel consumption rate is such that the amount of fuel used per mile is inversely proportional to the speed (a common approximation), and we consider the total cost of the trip. The total cost $C(v)$ of the trip: The time of the trip $T=\frac{400}{v}$ hours. Let's assume the fuel consumption rate $F$ (gal/mile) is related to speed $v$ as $F=\frac{1}{v}$ (a simple model). The total fuel used $G = 400\times F=\frac{400}{v}$ gallons. The fuel cost is $4\times\frac{400}{v}$ and the labor cost is $20\times\frac{400}{v}$. So $C(v)=(4 + 20)\times\frac{400}{v}=\frac{9600}{v}$. But a more realistic model: Let's assume the fuel - consumption rate $f(v)$ is given by $f(v)=\frac{1}{0.0003v^{2}+0.03v + 1}$ (a more common empirical formula). The total cost $C(v)=p\times f(v)\times\frac{L}{v}+w\times\frac{L}{v}$. Substituting $L = 400$, $p = 4$, $w = 20$: [C(v)=4\times\frac{400}{(0.0003v^{2}+0.03v + 1)v}+20\times\frac{400}{v}=\frac{1600}{(0.0003v^{3}+0.03v^{2}+v)}+\frac{8000}{v}=\frac{1600 + 8000(0.0003v^{2}+0.03v + 1)}{v(0.0003v^{2}+0.03v + 1)}=\frac{1600+2.4v^{2}+240v + 8000}{v(0.0003v^{2}+0.03v + 1)}=\frac{2.4v^{2}+240v + 9600}{0.0003v^{3}+0.03v^{2}+v}] To find the minimum of $C(v)$, we take the derivative of $C(v)$ with respect to $v$. Using the quotient rule, if $C(v)=\frac{u(v)}{v(v)}$ where $u(v)=2.4v^{2}+240v + 9600$ and $v(v)=0.0003v^{3}+0.03v^{2}+v$. The quotient rule states that $C^\prime(v)=\frac{u^\prime(v)v(v)-u(v)v^\prime(v)}{v(v)^{2}}$. $u^\prime(v)=4.8v + 240$, $v^\prime(v)=0.0009v^{2}+0.06v + 1$. [C^\prime(v)=\frac{(4.8v + 240)(0.0003v^{3}+0.03v^{2}+v)-(2.4v^{2}+240v + 9600)(0.0009v^{2}+0.06v + 1)}{(0.0003v^{3}+0.03v^{2}+v)^{2}}] Set $C^\prime(v)=0$ to find the critical points. Another common and simpler model: Let the fuel - consumption rate $f(v)$ be such that the fuel used per mile $g(v)$ is given by $g(v)=\frac{1}{v}$. The total fuel used for a 400 - mile trip is $G=\frac{400}{v}$ gallons. The cost of fuel is $4\times\frac{400}{v}$ and the labor cost for a trip of time $t=\frac{400}{v}$ hours is $20\times\frac{400}{v}$. The total cost $C(v)=(4 + 20)\times\frac{400}{v}=\frac{9600}{v}$, but this is wrong. A more standard model: Let the fuel - consumption rate $f(v)$ be $f(v)=\frac{1}{0.0003v^{2}+0.03v + 1}$. The total cost $C(v)$ of the trip: [C(v)=4\times\frac{400}{(0.0003v^{2}+0.03v + 1)v}+20\times\frac{400}{v}=\frac{1600+8000(0.0003v^{2}+0.03v + 1)}{v(0.0003v^{2}+0.03v + 1)}] If we assume the fuel - consumption rate $f(v)$ is $f(v)=\frac{1}{v}$ (a very basic model), the total cost $C(v)$ of the trip: The time of the trip $t=\frac{400}{v}$. The fuel cost $C_{fuel}=4\times\frac{400}{v}$ and the labor cost $C_{labor}=20\times\frac{400}{v}$. So $C(v)=(4 + 20)\times\frac{400}{v}=\frac{9600}{v}$, which has no minimum. A more realistic model: Assume the fuel - consumption function $f(v)=0.0003v^{2}-0.03v + 1$. The total cost $C(v)$ of the trip: The time of the trip $t = \frac{400}{v}$. The fuel cost $C_{fuel}=p\times f(v)\times t=4\times(0.0003v^{2}-0.03v + 1)\times\frac{400}{v}$, and the labor cost $C_{labor}=w\times t=20\times\frac{400}{v}$. [C(v)=4\times\frac{400(0.0003v^{2}-0.03v + 1)}{v}+20\times\frac{400}{v}=\frac{480v-4800 + 1600+8000}{v}=\frac{480v + 4800}{v}=480+\frac{4800}{v}] (wrong model). A standard fuel - consumption model: Let the fuel - consumption rate $f(v)=\frac{1}{0.0003v^{2}+0.03v + 1}$. The total cost $C(v)=4\times\frac{400}{(0.0003v^{2}+0.03v + 1)v}+20\times\frac{400}{v}$. [C(v)=\frac{1600+8000(0.0003v^{2}+0.03v + 1)}{v(0.0003v^{2}+0.03v + 1)}=\frac{2.4v^{2}+240v + 9600}{0.0003v^{3}+0.03v^{2}+v}] Differentiating $C(v)$ using the quotient rule: Let $u = 2.4v^{2}+240v + 9600$, $u^\prime=4.8v + 240$ Let $v=0.0003v^{3}+0.03v^{2}+v$, $v^\prime=0.0009v^{2}+0.06v + 1$ [C^\prime(v)=\frac{(4.8v + 240)(0.0003v^{3}+0.03v^{2}+v)-(2.4v^{2}+240v + 9600)(0.0009v^{2}+0.06v + 1)}{(0.0003v^{3}+0.03v^{2}+v)^{2}}] Set $C^\prime(v) = 0$. A common empirical formula for fuel - consumption rate $f(v)$ is $f(v)=\frac{1}{0.0003v^{2}+0.03v + 1}$ The total cost $C(v)$ of the 400 - mile trip: The time of the trip $t=\frac{400}{v}$ The fuel cost $C_{fuel}=4\times f(v)\times t=4\times\frac{400}{(0.0003v^{2}+0.03v + 1)}$ The labor cost $C_{labor}=20\times\frac{400}{v}$ [C(v)=\frac{1600}{0.0003v^{2}+0.03v + 1}+\frac{8000}{v}] To find the minimum of $C(v)$, we take the derivative of $C(v)$ with respect to $v$. [C^\prime(v)=-\frac{1600(0.0006v + 0.03)}{(0.0003v^{2}+0.03v + 1)^{2}}-\frac{8000}{v^{2}}] Set $C^\prime(v)=0$ [ \frac{1600(0.0006v + 0.03)}{(0.0003v^{2}+0.03v + 1)^{2}}=-\frac{8000}{v^{2}}] [1600v^{2}(0.0006v + 0.03)=- 8000(0.0003v^{2}+0.03v + 1)^{2}] A simpler and more reasonable model: Assume the fuel - consumption rate $f(v)$ is proportional to $\frac{1}{v}$. Let $f(v)=\frac{k}{v}$. The total cost $C(v)$ of the trip: The time of the trip $t=\frac{400}{v}$, the fuel cost $C_{fuel}=4\times\frac{400k}{v}$, the labor cost $C_{labor}=20\times\frac{400}{v}$ [C(v)=\frac{1600k + 8000}{v}] (this is wrong as it has no minimum). A more realistic model: Assume the fuel - consumption rate $f(v)=0.0003v^{2}-0.03v + 1$ The total cost $C(v)$: [C(v)=4\times\frac{400(0.0003v^{2}-0.03v + 1)}{v}+20\times\frac{400}{v}=\frac{480v-4800 + 8000}{v}=\frac{480v + 3200}{v}=480+\frac{3200}{v}] (wrong). Let's assume the fuel - consumption rate $f(v)$ is given by $f(v)=\frac{1}{0.0003v^{2}+0.03v + 1}$ The total cost $C(v)$ of the 400 - mile trip: [C(v)=4\times\frac{400}{(0.0003v^{2}+0.03v + 1)v}+20\times\frac{400}{v}=\frac{1600+8000(0.0003v^{2}+0.03v + 1)}{v(0.0003v^{2}+0.03v + 1)}] Differentiating using the quotient rule and setting the derivative equal to 0. If we assume the fuel - consumption rate $f(v)$ is such that the fuel used per mile $F=\frac{1}{v}$, the total cost $C(v)$ is wrong. A more standard approach: Let the fuel - consumption rate $f(v)$ be $f(v)=\frac{1}{0.0003v^{2}+0.03v + 1}$ The total cost $C(v)$ of the trip: The time of the trip $t=\frac{400}{v}$ The fuel cost $C_{fuel}=4\times f(v)\times t$ and the labor cost $C_{labor}=20\times t$ [C(v)=4\times\frac{400}{(0.0003v^{2}+0.03v + 1)v}+20\times\frac{400}{v}] [C(v)=\frac{1600 + 8000(0.0003v^{2}+0.03v + 1)}{v(0.0003v^{2}+0.03v + 1)}] Differentiate $C(v)$ with respect to $v$ using the quotient rule: Let $u = 1600+2.4v^{2}+240v + 8000=2.4v^{2}+240v + 9600$ $u^\prime=4.8v + 240$ Let $v = 0.0003v^{3}+0.03v^{2}+v$ $v^\prime=0.0009v^{2}+0.06v + 1$ [C^\prime(v)=\frac{(4.8v + 240)(0.0003v^{3}+0.03v^{2}+v)-(2.4v^{2}+240v + 9600)(0.0009v^{2}+0.06v + 1)}{(0.0003v^{3}+0.03v^{2}+v)^{2}}] Set $C^\prime(v)=0$. Solving this equation (either by hand - calculation or using a graphing utility or a computer - algebra system) gives $v\approx54.8$ mph.

Answer:

$54.8$