A manufacturer sells a certain article to dealers at a rate of $20 each if less than 50 are ordered. If 50 or more are ordered(up to 600) the price per article is reduced at a rate of 2 cents times the number ordered. What size order will produce the maximum amount of money for the manufacturer?
To find the order size that will produce the maximum amount of money for the manufacturer, we need to analyze the cost function.
Let's begin by breaking down the cost for the different order sizes.
For orders less than 50, the cost per article is $20 each. Therefore, the cost function for orders less than 50 can be represented as:
Cost_1(x) = 20x
Where x is the number of articles ordered.
For orders of 50 or more (up to 600), the price per article is reduced at a rate of 2 cents (or $0.02) times the number ordered. Therefore, the cost function for orders of 50 or more can be represented as:
Cost_2(x) = (20 - 0.02x)x
To find the order size that maximizes the manufacturer's profit, we need to compare the two cost functions.
To find the maximum amount of money for the manufacturer, we need to find the maximum value of the total revenue (sales revenue) minus the total cost.
Let's define the profit function as:
Profit(x) = Revenue(x) - Cost(x)
To find the maximum profit, we need to maximize this function.
Revenue(x) can be calculated by multiplying the selling price by the number of articles ordered:
Revenue(x) = Price(x) * x
For orders less than 50, the price per article is $20 each, resulting in:
Revenue_1(x) = 20x
For orders of 50 or more (up to 600), the price per article is reduced at a rate of 2 cents (or $0.02) times the number ordered:
Revenue_2(x) = (20 - 0.02x)x
Since we are comparing the revenue for orders of less than 50 and orders of 50 or more, the profit function can be expressed as:
Profit(x) = [(20x) - (20 - 0.02x)x]
To find the order size that maximizes profit, we need to find the critical points of the profit function where the derivative is zero.
Let's differentiate the profit function with respect to x:
Profit'(x) = [(20 - 0.02x) - (20 - 0.04x)] - [(20x) - (20 - 0.02x)]
Simplifying the expression:
Profit'(x) = 0.02x - 0.02x^2
To find the critical points, we need to set the derivative equal to zero and solve for x:
0.02x - 0.02x^2 = 0
Factoring out 0.02x:
0.02x(1 - x) = 0
Setting each factor equal to zero:
0.02x = 0 --> x = 0 (not applicable in this context)
1 - x = 0 --> x = 1
To determine if x = 1 is a maximum or minimum, we need to analyze the second derivative of the profit function.
To find the second derivative of the profit function, we differentiate the derivative once more:
Profit''(x) = 0.02 - 0.04x
Substituting x = 1 into the second derivative:
Profit''(1) = 0.02 - 0.04(1)
Profit''(1) = 0.02 - 0.04
Profit''(1) = -0.02
Since the second derivative is negative, x = 1 is a maximum point.
Therefore, an order size of 1 will produce the maximum amount of money for the manufacturer.
To determine the order size that will produce the maximum amount of money for the manufacturer, we need to find the point at which the revenue is maximized.
Let's break down the problem step by step:
Step 1: Calculate the revenue for an order of less than 50 articles.
For orders less than 50, the price per article is fixed at $20. Therefore, the revenue for each order will be:
Revenue = Price per article × Number of articles
Revenue = $20 × Number of articles
Step 2: Calculate the revenue for an order of 50 or more articles.
For orders of 50 or more, the price per article is reduced by 2 cents times the number of articles. Let's represent the number of articles in an order as "x". Therefore, the price per article for an order of 50 or more will be:
Price per article = $20 - (0.02 * x)
The revenue for an order of 50 or more articles will be:
Revenue = (Price per article) × Number of articles
Revenue = ($20 - 0.02x) × x
Step 3: Determine the equation for the total revenue.
The total revenue will be the sum of the revenues from both steps, as long as it is within the given range of 1 to 600 articles:
Total Revenue = $20 × Number of articles for orders less than 50
+ ($20 - 0.02x) × x for orders of 50 or more
Step 4: Simplify the equation.
Total Revenue = 20x (1 ≤ x < 50)
= (20x - 0.02x^2) (50 ≤ x ≤ 600)
Step 5: Find the order size that maximizes the total revenue.
To find the order size that maximizes the total revenue, differentiate the equation with respect to "x" and set it equal to zero. Then solve for "x".
d(Total Revenue)/dx = 20 - 0.04x = 0
0.04x = 20
x = 500
So, an order size of 500 articles will produce the maximum amount of money for the manufacturer.