Instruction: You can hand-write or type your answers and turn them in by due dat
ID: 3114300 • Letter: I
Question
Instruction: You can hand-write or type your answers and turn them in by due date. You can work individually or in a group of 2-3 students. Only one submission is needed for each group. 1. The profit P (in dollars) from selling x laptop computers is given by P =-0.04x2 + 25x-1500. a. Find the additional profit when the sales increase from150 to 151 units. b. Find the marginal profit when = 150 c. Compare the results of part (a) and (b). 2. Use the information in the table to find the models and answer the question below Price (p) Total revenue (TR) Marginal revenue (MR) Quantity produced and sold (Q 160 130 90 50 10 600 80 10 60 600 Use graphing utility (or your graphing calculator) to find the quadratic model that relates the total revenue (TR) to the quantity produced and sold (Q) Using derivatives, find a model for marginal revenue from the model you found in part a. b. Calculate the marginal revenue for all values of Q using your model in part (b), and compare these values with the actual values given. How good is your model? c.Explanation / Answer
1)
Given that profit (P) in selling x laptops is given by:
P = -0.04x2 + 25x -1500
a) Addtional profit when sales increase from 150 to 151 units
= P/x = [ P(151) - P(150) ] / (151 - 150)
= (1362.96 - 13500 / 1
= 12.96 dollars
b) To find Marginal profit when x = 150, first we will find the derivative of the profit function P(x) :
so, dP/dx = P'(x) = -0.08x + 25
so, marginal profit at x =150 is : P'(150) = -0.08*150 + 25
= -12 + 25 = 13 dollars
c) Results of (a) and (b) are approximately same.
2)
By using the given table and info therein,
a) To find the Total Revenue function or R(x), first we will generate the demand function for the data given. The demand function is always linear and is of the form : p = D(x), where p is price and x is quantity.
In linear form it can be written as p = mx + b,
here we first have to find the values of m and b from the given data of price and quantity. The given data can be written in ordered pairs form of (x, p) like : (0, 0), (2, 140), (4, 120) and so on.
Now to find the value of m which is the slope of this linear equation, we need to calculate it by using formula :
m = (y2 – y1) / (x2 – x1), which gives us ==> m = (120 – 140) / 4 – 2 ==> m = -10
Now with the value of m, we can rewrite our demand function as :
p = -10x + b
Now further to find value of b, we can substitute any one of the ordered pairs from the data above into this equation to get,
120 = -10*4 + b giving b = 160
So, finally the demand function is : p = -10x + 160
Revenue function : the revenue function is found by simply observing that Revenue = price * quantity
So in our case, R(x) = p*x = (-10x + 160)*x
or R(x) = -10x2 + 160x is the required Revenue function in quadratic form.
b)
Marginal revenue : It is the derivative of the Revenue function, so we will take the derivative of R(x) to find marginal revenue,
We have R(x) = -10x^2 + 160x
So, (d/dx)R(x) = R’(x) = -20x + 160 is the required Marginal revenue function.
c)
Now Marginal revenue in each case is :
For x = 0, Marginal revenue = NA
For x = 2, Marginal revenue = (using model = $120, actual = $130)
For x = 4, Marginal revenue = (using model = $80, actual = $90)
For x = 6, Marginal revenue = (using model = $40, actual = $50)
For x = 8, Marginal revenue = (using model = $0, actual = $10)
For x = 10, Marginal revenue = (using model = (-) $40, actual = (-) $30)
Model is giving flat $10 less marginal revenue than the actual.