A car filled with x kg of gasoline consumes [tex]\displaystyle \large{\frac{100+x}{100}e^{kv}}[/tex] 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:

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

Step-by-step explanation:

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

[tex]\displaystyle \frac{dx}{dt} = -\frac{100 + x}{100}\, e^{k\, v}[/tex].

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 [tex]x[/tex] should be negative.

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

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

[tex]\displaystyle \frac{dx}{100 + x} = -\frac{e^{k\, v}\, dt}{100}[/tex].

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

[tex]\displaystyle \int \frac{dx}{100 + x} = -\frac{e^{k\, v}}{100}\int dt[/tex].

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

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

[tex]\displaystyle \ln | 100 + x_{0} | = C[/tex].

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

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

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

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

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

[tex]\displaystyle \ln | 100 | = -\frac{e^{k\, v}}{v} + \ln | 100 + x_{0}|[/tex].

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

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

[tex]\displaystyle \ln | 100 + x_{0}| = \ln | 100 | + \frac{e^{k\, v}}{v}}[/tex].

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

Differentiate the right hand side with respect to [tex]v[/tex]:

[tex]\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}[/tex].

Setting this first derivative to [tex]0[/tex] and solving for [tex]v[/tex] gives:

[tex]k\,v - 1 = 0[/tex].

[tex]v = (1/k)[/tex].

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

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

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

[tex]\displaystyle \ln | 100 + x_{0}| = \ln | 100 | + k\, e[/tex].

[tex]e^{\ln| 100 + x_{0}|} = e^{\ln|100| + k\, e}[/tex].

[tex]e^{\ln| 100 + x_{0}|} = e^{\ln|100|}\, e^{k\, e}[/tex].

[tex]100 + x_{0} = 100\, e^{k\, e}[/tex].

[tex]x_{0} = 100\, e^{k\, e} - 100[/tex].

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

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