Question

A car filled with x kg of gasoline consumes 7B100%7De%5E%7Bkv%7D%7D" id="TexFormula1" title="\displaystyle \large{\frac{100+x}{100}e^{kv}}" alt="\displaystyle \large{\frac{100+x}{100}e^{kv}}" align="absmiddle" class="latex-formula"> kg/hr of gasoline when driving at v km/hr. (k is a positive constant.) Given that this car drives to a destination 100 km away at a constant speed, find the initial amount of gasoline and the driving speed such that gasoline consumption is minimized. Assume that the car stops immediately after running out of gasoline.

Answer 1

Answer:

The vehicle should start with $(100\, e^{ke} - 100)\; {\rm kg}$ of fuel and drive at a speed of $(1/k)\; {\rm km \cdot hr^{-1}}$.

Step-by-step explanation:

Let $t\; {\rm hr}$ denote the number of hours after the vehicle started. As in the question, let the amount of fuel currently on this vehicle be $x\; {\rm kg}$. The question states that the vehicle consumes fuel at a rate ($dx/dt$) of $((100 + x) / 100)\, e^{kv}$. In other words:

$\displaystyle \frac{dx}{dt} = -\frac{100 + x}{100}\, e^{k\, v}$.

Note the minus sign in front of the right-hand side. The amount of fuel on this vehicle decreases over time. Hence, the rate of change in $x$ should be negative.

This equation is a separable ordinary differential equation. The variables are $x$ and $t$. Solve this ODE to find an expression of $x\!$ (fuel in the vehicle) in terms of $t\!$ (time.) Follow these steps:

Rearrange this equation such that all $x$ and $dx$ are are on the same side of the equation, while $t$ and $dt$ on all on the other side.

$\displaystyle \frac{dx}{100 + x} = -\frac{e^{k\, v}\, dt}{100}$.

Integrate both sides, and the equality should still hold. Note that $k$ and $v$ are considered as constants. Be sure to include the constant of integration $C$ on one side of the equation.

$\displaystyle \int \frac{dx}{100 + x} = -\frac{e^{k\, v}}{100}\int dt$.

$\displaystyle \ln | 100 + x | = -\frac{(e^{k\, v})\, t}{100} + C$.

Let $x_{0}$ denote the initial amount of fuel on this vehicle (i.e., the value of $x$ when $t = 0$). The constant of integration $C$ should ensure that $x = x_{0}$ when $t = 0$. Thus:

$\displaystyle \ln | 100 + x_{0} | = C$.

Hence, the value of the constant of integration should be $\ln | 100 + x_{0} |$. Therefore:

$\displaystyle \ln | 100 + x | = -\frac{(e^{k\, v})\, t}{100} + \ln | 100 + x_{0}|$.

Since the speed of the vehicle is constant at $v\; {\rm km\cdot hr^{-1}}$, the time required to travel $100\; {\rm km}$ would be $(100 / v)\; {\rm hr}$.

For optimal use of the fuel, the vehicle should have exactly $x = 0$ fuel when the destination is reached. Therefore, $x = 0$ at $t = 100 / v$. Hence:

$\displaystyle \ln | 100 | = -\frac{(e^{k\, v})\, (100 / v)}{100} + \ln | 100 + x_{0}|$.

$\displaystyle \ln | 100 | = -\frac{e^{k\, v}}{v} + \ln | 100 + x_{0}|$.

Notice that $\ln|100 + x_{0}|$ is monotone increasing with respect to $x_{0}$ as long as $100 + x_{0} > 0$. Thus, given that $x_{0} > 0$, $x_{0}\!$ would be minimized if and only if the surrogate $\ln|100 + x_{0}|\!$ is minimized.

While the goal is to find the $v$ that minimize $x_{0}\!$, finding the $v\!$ that minimizes $\ln|100 + x_{0}|$ would achieve the same purpose.

$\displaystyle \ln | 100 + x_{0}| = \ln | 100 | + \frac{e^{k\, v}}{v}}$.

The $\texttt{RHS}$ of this equation is indeed convex with respect to $v$ ($v > 0$.) Thus, the $\texttt{RHS}\!$ could be minimized by setting the first derivative with respect to $v$ to $0$.

Differentiate the right hand side with respect to $v$:

$\begin{aligned} & \frac{d}{dv}\left[\frac{e^{k\, v}}{v}\right] \\ =\; & \frac{k\, e^{k\, v}}{v} - \frac{e^{k\, v}}{v^{2}}\\ =\; & \frac{(k\, v - 1)\, e^{k\, v}}{v^{2}}\end{aligned}$.

Setting this first derivative to $0$ and solving for $v$ gives:

$k\,v - 1 = 0$.

$v = (1/k)$.

Therefore, the amount of fuel required for this trip is minimized when $v = (1/k)\; {\rm km \cdot hr^{-1}}$.

Substitute $v$ back and solve for $x_{0}$ (initial amount of fuel on the vehicle.)

$\displaystyle \ln | 100 + x_{0}| = \ln | 100 | + \frac{e^{k\, (1/k)}}{(1/k)}}$.

$\displaystyle \ln | 100 + x_{0}| = \ln | 100 | + k\, e$.

$e^{\ln| 100 + x_{0}|} = e^{\ln|100| + k\, e}$.

$e^{\ln| 100 + x_{0}|} = e^{\ln|100|}\, e^{k\, e}$.

$100 + x_{0} = 100\, e^{k\, e}$.

$x_{0} = 100\, e^{k\, e} - 100$.

In other words, the initial amount of fuel on the vehicle should be $(100\, e^{k\, e} - 100)\; {\rm kg}$.