Question

A car filled with x kg of gasoline consumes
`\displaystyle \large{(100+x)/(100)e^(kv)}` 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

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)}`.

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 (dx)/(dt) = -(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 (dx)/(100 + x) = -(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 (dx)/(100 + x) = -(e^(k\, v))/(100)\int dt`.

`\displaystyle \ln | 100 + x | = -((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 | = -((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 | = -((e^(k\, v))\, (100 / v))/(100) + \ln | 100 + x_(0)|`.

`\displaystyle \ln | 100 | = -(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 | + (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} & (d)/(dv)\left[(e^(k\, v))/(v)\right] \\ =\; & (k\, e^(k\, v))/(v) - (e^(k\, v))/(v^(2))\\ =\; & ((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 | + (e^(k\, (1/k)))/((1/k))}`.

`\displaystyle \ln | 100 + x_(0)| = \ln | 100 | + k\, e`.

`e^\ln = e^(\ln|100| + k\, e)`.

`e^ 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}`.