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

1. Define the cost - function:
   • Let the speed of the vehicle be (v) mph. The time (t) taken for a (400) - mile trip is (t=\frac{400}{v}) hours.
   • The fuel cost per gallon is (p = 4) dollars per gallon, and the fuel consumption rate (w) is (20) dollars per hour.
   • The total cost (C) of the trip is the sum of the fuel cost and the time - related cost. The amount of fuel used for the trip is related to the time of the trip. If we assume the fuel consumption rate in terms of gallons per hour is some function of speed, but here we can also consider the total cost in terms of the given variables. The total cost (C) of the trip is the cost of fuel and the cost of time. The cost of time is (20t) (since the cost per hour is (20) dollars) and the cost of fuel is related to the distance and fuel - efficiency. However, we can also set up the cost function as (C(v)=4\times\frac{400}{v}\times k+20\times\frac{400}{v}), where (k) is a factor related to fuel consumption per mile. In a more straightforward way, we know that the total cost (C(v)) of the (400) - mile trip is the sum of the cost due to fuel and the cost due to time. The time taken for the trip is (t=\frac{400}{v}) hours. The cost function (C(v)) is (C(v)=4\times\frac{400}{v}\times1 + 20\times\frac{400}{v}) (assuming a simple model where the fuel cost is based on the amount of fuel used during the trip and the time cost is based on the time of the trip).
   • Simplify the cost function:
     • (C(v)=(4 + 20)\times\frac{400}{v}=24\times\frac{400}{v}=\frac{9600}{v}).
2. Find the minimum of the cost - function:
   • To find the minimum of the function (y = C(v)=\frac{9600}{v}=9600v^{-1}), we take the first - derivative with respect to (v).
     • Using the power rule (\frac{d}{dv}(x^n)=nx^{n - 1}), we have (C^\prime(v)=-9600v^{-2}=-\frac{9600}{v^{2}}).
     • Since (C^\prime(v)) is never zero for (v\gt0) (the speed (v) must be positive), and we know that the function (C(v)=\frac{9600}{v}) is a hyperbola. As (v) increases, (C(v)) decreases. In a more practical sense, if we consider the domain of non - negative real numbers for speed, and assume no other constraints, the cost function (C(v)) is a decreasing function for (v\gt0). But if we consider real - world constraints (e.g., speed limits, vehicle performance), we assume we want to find the minimum cost within a reasonable speed range. If we assume no upper - bound on speed for the sake of this mathematical model, as (v) gets larger, the cost per mile (in terms of time and fuel combined) decreases. However, if we consider the physical meaning, we know that the cost function (C(v)) is a hyperbola (y=\frac{k}{v}) ((k = 9600) in our case).
     • In a more general calculus approach, we can also consider the second - derivative (C^{\prime\prime}(v)=\frac{19200}{v^{3}}\gt0) for (v\gt0), which means the function is concave up for (v\gt0).
     • To minimize the cost, we note that as (v) increases, (C(v)) decreases. But if we assume we are looking for a non - zero and non - infinite speed, and we consider the fact that the cost function (C(v)) is of the form (y=\frac{k}{v}), we can say that in the absence of other constraints, we want to make (v) as large as possible. But if we consider real - world scenarios, we assume we want to find the minimum within a reasonable range. If we assume we are talking about a normal driving situation, we can also use the fact that the cost function (C(v)) is a simple rational function.
     • Let's assume we want to find the minimum of the cost function by setting the first - derivative equal to zero (even though in this simple form, the first - derivative is never zero for non - zero (v)). In a more practical sense, we can rewrite the cost function as (C(v)=\frac{9600}{v}), and we know that this is a hyperbola. The minimum cost occurs when (v) is maximized within the legal and safe speed range. If we assume no such constraints for the pure mathematical model, we can say that the function (C(v)) is a decreasing function for (v\in(0,\infty)).
     • If we consider the fact that we want to minimize (C(v)) and (C(v)=\frac{9600}{v}), we know that as (v) gets larger, (C(v)) gets smaller. But in a real - world context, if we assume a speed limit of, say, (80) mph (a common highway speed limit in some areas), we can say that within the legal speed range, we want to drive at the maximum legal speed.
     • Mathematically, if we consider the domain of positive real numbers for (v), the function (C(v)) has no minimum in the strict sense (it approaches (0) as (v\to\infty)). But if we assume a practical upper - bound (v_{max}), the minimum cost in the interval ((0,v_{max}]) occurs at (v = v_{max}). Let's assume a speed limit of (80) mph.

Explanation:

Step1: Define the cost function

The time (t) for a (400) - mile trip at speed (v) is (t=\frac{400}{v}). The total cost (C(v)) of the trip is the sum of fuel cost and time cost. The fuel cost is (4\times\frac{400}{v}) and the time cost is (20\times\frac{400}{v}), so (C(v)=(4 + 20)\times\frac{400}{v}=\frac{9600}{v}).

Step2: Analyze the function for minimum

Take the first - derivative (C^\prime(v)=-\frac{9600}{v^{2}}). Since (C^\prime(v)\lt0) for (v\gt0), the function (C(v)) is a decreasing function on the interval ((0,\infty)). In a practical sense, within the legal speed limit (assume (80) mph), the minimum cost occurs at the maximum speed in that range.

Answer:

80 (assuming a speed limit of 80 mph. If no speed - limit is given in the problem, the mathematical model suggests that the cost decreases as the speed increases, but in real - world, there are speed limits. Here we assume a common speed limit value for illustration purposes)