|Type of paper:||Term paper|
|Categories:||Financial analysis Microeconomics|
Assignment: Determine the maximum revenue for the given companies Maximizing Revenue ---- Part I
The marketing department at Gadgets Galore has found that when certain iPods are sold at a price of p dollars per unit, the revenue R (in dollars) as a function of the price p can be demonstrated by the function given in the graph below.
Use the above graph to answer the questions stated in Part I of the Response Sheet.
PART I: Maximizing Revenue (Quantitative Literacy)
1. Interpretation: Using complete sentences, explain the information represented by the graph.
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The vertical axis of the graph demonstrates the revenue(R) while the horizontal axis represents the price (P). The revenue and the price are correlated. The different prices set for the iPod determine the ranging revenue that will be earned. Therefore, the rates of revenue depends on the set price. The graph shows the price of the iPods from $0-$140 and the revenue gained from $0-$750000. $750000 is the maximum revenue while $140 is the maximum price of the product.
2. Representation and Calculations: Find the approximate revenue when the price of the item is...
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p=$20 R= $360000
p=$40 R= $600000
p=$60 R= $710000
p=$80 R= $710000
p=$100 R= $600000
p=$120 R= $360000
3. Application and analysis: Make a judgment and draw appropriate conclusions based on the quantitative analysis of the data represented in the given graph. (Hint: Does the data lead you to the maximum point on the graph? How does knowing this projected maximum point help this company?) Write at least two or three sentences to explain your answer.
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The graph demonstrates the dependence of revenue(r) and price (p). If Gadgets Galore exceed the maximum price to sell the iPods, the maximum revenue decreases. Therefore, the firm has to sell the products at a price of $70 each to achieve a maximum revenue of $750000.
4. Assumptions: What assumptions did you make in your estimate of the maximum revenue and the needed price to charge to reach this maximum revenue? (Hint: When looking at the graph, what did you assume about the data shown?) Write at least two or three sentences to explain your answer.
The first assumption is that the two axes, price and revenue, are variably correlated. The second assumption is that once the iPod is sold at a price of more than $70, the maximum revenue begins to drop. Another assumption is that the points on the graph are linear and it is symmetrical from its maximum point.
PART II: Maximizing Revenue (Communication and Critical Thinking - includes QL6)
Write a 250 - 400 word essay (about 3 - 4 paragraphs) explaining how to solve the following problem:
The John Deere Company has found that the revenue from sales of heavy-duty tractors is a function of the unit price p that it charges. If the revenue R is R(p) = - 0.5p2 + 1900p , what unit price p should be charged to maximize revenue? What is the maximum revenue? What recommendations would you give the John Deere Company?
Pretend you are explaining this to someone who has not taken a math class in several years. Be sure to include the following in your explanation:
what the p and R represent,
the actual value of the maximum revenue and the price needed to achieve this value, and
recommendations you would give the John Deere Company (Hint: How can the data you calculated be used to benefit the company? For example, why shouldn't the company charge $2000 or more for each tractor?) Use proper English and full sentences with ideas separated into paragraphs. Spelling and grammar count!
In the equation R(p) = - 0.5p2 + 1900p, P in the equation represents the price that each tractor was sold while the R stands for the revenue gained by the company after selling a given number of tractors. The result of R is revenue for a price that the tractors are sold for. The revenue of the sales increase to a maximum point over a given price but drops if the price at the maximum point exceeds. The maximum revenue determines the maximum price that the tractors will be sold for.
Maximum revenue can be figured by two methods:
The revenue, R(p), is on the y axis while the price, p, is on the x axis. Plot the figures of the revenue and the price against each other until the gradient reads zero and starts to drop. The maximum revenue is the point on y axis where the gradient reads zero while the maximum price to charge a product is where the gradient reads zero from the x axis.
In this formula, the function is equated to zero to solve for the price when maximum revenue is achieved. The result of price (p) is then substituted in the equation to calculate the figure for the maximum revenue. Below is how the derivative method is used to find a solution for the maximum revenue:
R(p) = -0.5p2 + 1900pThe first derivative of R (p) is zero:
dRpdp=ddp -0.5p2+ 1900p=02(-0.5p)+1900=0-p+1900=0-p=-1900Divide the two sides of the equation by -1:
-p-1=-1900-1p=$1900Therefore, the maximum price to sell a tractor is $1900.
Solve the maximum revenue by replacing the value of p in the main formula:
Rp= -0.5p2+ 1900pR(p) = -0.519002 + 19001900=-0.5(3610000)+3610000=-1805000+3610000=$180500If the firm charges $2000 or every tractor, the revenue would be:
R(p) = -0.5p2 + 1900pR(p) = -0.520002 + 1900(2000)=-2000000+3800000=$180000The revenue drops when the company charges $2000 for every tractor. Therefore, is best if the company sells its products at the maximum price of $1900. If otherwise, the maximum revenue would decrease.
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