d) What price will maximize the profit?

Managerial Economics: Applications, Strategies and Tactics (MindTap Course List)
14th Edition
ISBN:9781305506381
Author:James R. McGuigan, R. Charles Moyer, Frederick H.deB. Harris
Publisher:James R. McGuigan, R. Charles Moyer, Frederick H.deB. Harris
Chapter8: Cost Analysis
Section: Chapter Questions
Problem 5E
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Solve part d only

Solution of ist 3 parts is provided 

Only d 

Partd only

a)
The break-even points refers to the points
at which the company's revenue equals
its costs. We can find the break-even
points by setting the revenue function
equal to the cost function and solving for
X.
The revenue function is given by:
3
R(x)=1250x-(=)x²
5
The cost function is given by:
2
C(x) = 28,000+ (²)x² +222x
Now putting these two function equal to
each other
2
1250x-()x² = 28,000+ (²)x² +222x
1250x = 28,000+x²+222x
x²-1028x+28,000 = 0
by solving we get,
X = 28
Thus breakeven point = 28 units
Step 3: Find the maximum revenue
b) The maximum revenue refers to the
maximum amount of money that the firm
can make by selling its product. We can
find the maximum revenue by setting the
derivative of the revenue function equal
to zero and solving for x.
The derivative of the revenue function is
given by:
R'(x) = 1250 (6/5)x
Setting this equal to zero and solving for
x, we get:
x= 1041.67 ≈ 1042
Putting the value of x in revenue finction
we get,
R(x) = $651,041.60
Therefore, the maximum revenue is
$651,041.60, which occurs when the
company sells 1042 units.
Step 4: Form the required function
The profit function can be find out by the
difference between the revenue function
and the cost function. It is given by:
P(x) = R(x) - C(x)
Subbing the revenue and cost functions,
we get:
3
2
P(x)= 1250x – (²)x² – [28,000 + (²)x²
-)x² +1028x-28,000
P(x)=
To find the maximum profit, we can
involve similar technique as we
accomplished for tracking down the
maximum revenue. Setting the derivative
of the profit function equivalent to zero
and solving for x, we get:
2
(²)x + 1028 = 0
P'(x):
= -
Solving for x, we get:
x = 2570
Putting the value of x in profit function
we get,
Max profit = $1,292,980
Therefore, the maximum profit is
$1,292,80, which occurs when the
company sells 2570 units.
Solution
a) Thus breakeven point = 28 unit
b) the maximum revenue is $651,041.60,
which occurs when the company sells
1042 units.
c) the maximum profit is $1,292,80, which
occurs when the company sells 2570
units.
Transcribed Image Text:a) The break-even points refers to the points at which the company's revenue equals its costs. We can find the break-even points by setting the revenue function equal to the cost function and solving for X. The revenue function is given by: 3 R(x)=1250x-(=)x² 5 The cost function is given by: 2 C(x) = 28,000+ (²)x² +222x Now putting these two function equal to each other 2 1250x-()x² = 28,000+ (²)x² +222x 1250x = 28,000+x²+222x x²-1028x+28,000 = 0 by solving we get, X = 28 Thus breakeven point = 28 units Step 3: Find the maximum revenue b) The maximum revenue refers to the maximum amount of money that the firm can make by selling its product. We can find the maximum revenue by setting the derivative of the revenue function equal to zero and solving for x. The derivative of the revenue function is given by: R'(x) = 1250 (6/5)x Setting this equal to zero and solving for x, we get: x= 1041.67 ≈ 1042 Putting the value of x in revenue finction we get, R(x) = $651,041.60 Therefore, the maximum revenue is $651,041.60, which occurs when the company sells 1042 units. Step 4: Form the required function The profit function can be find out by the difference between the revenue function and the cost function. It is given by: P(x) = R(x) - C(x) Subbing the revenue and cost functions, we get: 3 2 P(x)= 1250x – (²)x² – [28,000 + (²)x² -)x² +1028x-28,000 P(x)= To find the maximum profit, we can involve similar technique as we accomplished for tracking down the maximum revenue. Setting the derivative of the profit function equivalent to zero and solving for x, we get: 2 (²)x + 1028 = 0 P'(x): = - Solving for x, we get: x = 2570 Putting the value of x in profit function we get, Max profit = $1,292,980 Therefore, the maximum profit is $1,292,80, which occurs when the company sells 2570 units. Solution a) Thus breakeven point = 28 unit b) the maximum revenue is $651,041.60, which occurs when the company sells 1042 units. c) the maximum profit is $1,292,80, which occurs when the company sells 2570 units.
+222
Suppose a company has fixed costs of $28,000 and variable cost per unit of
dollars, where x is the total number of units produced. Suppose further that the selling
price of its product is 1250 dollars per unit.
-
a) Find the break-even points.
b) Find the maximum revenue.
c) Form the profit function from the cost and revenue function and find maximum profit.
d) What price will maximize the profit?
Transcribed Image Text:+222 Suppose a company has fixed costs of $28,000 and variable cost per unit of dollars, where x is the total number of units produced. Suppose further that the selling price of its product is 1250 dollars per unit. - a) Find the break-even points. b) Find the maximum revenue. c) Form the profit function from the cost and revenue function and find maximum profit. d) What price will maximize the profit?
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