A person operates a delivery service in a southern city. His start-up costs amounted to $2500. He estimates that it costs him (in terms of gasoline, wear and tear on his car, etc.) $4.00 per delivery. He charges $6.50 per delivery. Let x represent the number of deliveries he makes.
(a) Express the cost C as a function of x, where x represents the quantity of items as given.
(b) Express the revenue R as a function of x.
(c) Determine the value of x for which revenue equals cost.
(d) Graph the equations y=C(x) and y=R(x) on the same axes, and interpret the graph.
(a) The cost C as a function of x is given by:
C(x) = 2500 + 4x
The initial start-up cost is $2500 and it costs him $4 per delivery.
(b) The revenue R as a function of x is given by:
R(x) = 6.50x
He charges $6.50 per delivery, so the revenue is simply the price per delivery multiplied by the number of deliveries.
(c) To determine the value of x for which revenue equals cost, we need to set C(x) equal to R(x) and solve for x:
2500 + 4x = 6.50x
2500 = 2.50x
x = 2500/2.50
x = 1000
So when he makes 1000 deliveries, the revenue will equal the cost.
(d) The graph of the equations y = C(x) and y = R(x) on the same axes will have x as the independent variable on the x-axis and the cost and revenue on the y-axis. The graph will be a linear function with a positive slope.
The graph will start at C = 2500 on the y-axis at x = 0, indicating the initial start-up cost. As the number of deliveries (x) increases, the cost (y) will increase at a rate of $4 per delivery. The revenue (y) will increase at a rate of $6.50 per delivery, starting from the origin.
The point of intersection of the two graphs represents the number of deliveries (x) at which the revenue equals the cost. In this case, it is x = 1000.
(a) The cost C as a function of x can be expressed as:
C(x) = $2500 + ($4.00 * x)
(b) The revenue R as a function of x can be expressed as:
R(x) = $6.50 * x
(c) To determine the value of x for which revenue equals cost, we can set C(x) equal to R(x) and solve for x:
$2500 + ($4.00 * x) = $6.50 * x
Simplifying the equation:
$2500 = $6.50 * x - $4.00 * x
$2500 = $2.50 * x
Dividing both sides by $2.50:
x = $2500 / $2.50
x = 1000
So, the value of x for which revenue equals cost is 1000 deliveries.
(d) To graph the equations y = C(x) and y = R(x) on the same axes, we can plot points for different values of x and calculate the corresponding values of C(x) and R(x).
When x = 0:
C(0) = $2500
R(0) = $0
When x = 500:
C(500) = $2500 + ($4.00 * 500) = $2500 + $2000 = $4500
R(500) = $6.50 * 500 = $3250
When x = 1000:
C(1000) = $2500 + ($4.00 * 1000) = $2500 + $4000 = $6500
R(1000) = $6.50 * 1000 = $6500
Plotting these points on a graph, we have:
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R(x) |\
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0 500 1000
The graph shows that initially, the cost C(x) is higher than the revenue R(x). As the number of deliveries increases, the revenue eventually surpasses the cost. This indicates that after a certain point (in this case, approximately 1000 deliveries), the delivery service starts generating profit.
To solve this problem, let's break it down step by step.
(a) Express the cost C as a function of x:
The cost for each delivery is $4.00, and the number of deliveries is represented by x. Therefore, the total cost can be expressed as C(x) = 4x.
(b) Express the revenue R as a function of x:
The person charges $6.50 per delivery, and the number of deliveries is represented by x. Therefore, the total revenue can be expressed as R(x) = 6.50x.
(c) Determine the value of x for which revenue equals cost:
To determine when the revenue equals the cost, we can set the cost function equal to the revenue function and solve for x:
4x = 6.50x
Simplifying, we have:
6.50x - 4x = 0
2.50x = 0
x = 0
Therefore, the value of x for which revenue equals cost is x = 0.
(d) Graph the equations y = C(x) and y = R(x) on the same axes, and interpret the graph:
To graph the equations, we can plot points for different values of x. Since the cost and revenue functions are linear, we only need two points for each function.
For example, let's choose x = 0, 2, and 4:
For x = 0, C(0) = 4(0) = 0 and R(0) = 6.50(0) = 0.
For x = 2, C(2) = 4(2) = 8 and R(2) = 6.50(2) = 13.
For x = 4, C(4) = 4(4) = 16 and R(4) = 6.50(4) = 26.
Now let's plot these points on a graph with x on the horizontal (x-axis) and the cost/revenue on the vertical (y-axis). We will plot C(x) as one line and R(x) as another.
The graph will show a linear increase for the revenue line (y = R(x)) since it goes up by $6.50 for every delivery, while the cost line (y = C(x)) will also increase linearly but at a slower rate since it only goes up by $4.00 for every delivery.
Interpretation of the graph:
The graph will show that as the number of deliveries (x) increases, both the cost and revenue increase. However, the revenue line will have a steeper slope compared to the cost line, indicating that the revenue grows faster than the cost.
At x = 0, both cost and revenue are zero since no deliveries are made. As x increases, the revenue surpasses the cost, indicating potential profit for the delivery service. The intersection point of the two lines represents the breakeven point, where the revenue equals the cost.