Answer:
The vehicle should start with 
of fuel and drive at a speed of 
.
Step-by-step explanation:
Let 
denote the number of hours after the vehicle started. As in the question, let the amount of fuel currently on this vehicle be 
. The question states that the vehicle consumes fuel at a rate (
) of 
. In other words:
.
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 
should be negative.
This equation is a separable ordinary differential equation. The variables are and 
. Solve this ODE to find an expression of 
(fuel in the vehicle) in terms of 
(time.) Follow these steps:
Rearrange this equation such that all and 
are are on the same side of the equation, while and 
on all on the other side.
.
Integrate both sides, and the equality should still hold. Note that 
and 
are considered as constants. Be sure to include the constant of integration 
on one side of the equation.
.
.
Let 
denote the initial amount of fuel on this vehicle (i.e., the value of when 
). The constant of integration should ensure that 
when . Thus:
.
Hence, the value of the constant of integration should be 
. Therefore:
.
Since the speed of the vehicle is constant at 
, the time required to travel 
would be 
.
For optimal use of the fuel, the vehicle should have exactly 
fuel when the destination is reached. Therefore, at 
. Hence:
.
.
Notice that 
is monotone increasing with respect to as long as 
. Thus, given that 
, 
would be minimized if and only if the surrogate 
is minimized.
While the goal is to find the that minimize , finding the 
that minimizes would achieve the same purpose.
.
The 
of this equation is indeed convex with respect to (
.) Thus, the 
could be minimized by setting the first derivative with respect to to 
.
Differentiate the right hand side with respect to :
![\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}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%20%26%20%5Cfrac%7Bd%7D%7Bdv%7D%5Cleft%5B%5Cfrac%7Be%5E%7Bk%5C%2C%20v%7D%7D%7Bv%7D%5Cright%5D%20%5C%5C%20%3D%5C%3B%20%26%20%5Cfrac%7Bk%5C%2C%20e%5E%7Bk%5C%2C%20v%7D%7D%7Bv%7D%20-%20%5Cfrac%7Be%5E%7Bk%5C%2C%20v%7D%7D%7Bv%5E%7B2%7D%7D%5C%5C%20%3D%5C%3B%20%26%20%5Cfrac%7B%28k%5C%2C%20v%20-%201%29%5C%2C%20e%5E%7Bk%5C%2C%20v%7D%7D%7Bv%5E%7B2%7D%7D%5Cend%7Baligned%7D)
.
Setting this first derivative to and solving for gives:
.
.
Therefore, the amount of fuel required for this trip is minimized when 
.
Substitute back and solve for (initial amount of fuel on the vehicle.)
.
.
.
.
.
.
In other words, the initial amount of fuel on the vehicle should be 
.
