Solve part d onlySolution of ist 3 parts is provided Only d Partd only a)The break-even points refers to the pointsat which the company's revenue equalsits costs. We can find the break-evenpoints by setting the revenue functionequal to the cost function and solving forX.The revenue function is given by:3R(x)=1250x-(=)x²5The cost function is given by:2C(x) = 28,000+ (²)x² +222xNow putting these two function equal toeach other21250x-()x² = 28,000+ (²)x² +222x1250x = 28,000+x²+222xx²-1028x+28,000 = 0by solving we get,X = 28Thus breakeven point = 28 unitsStep 3: Find the maximum revenueb) The maximum revenue refers to themaximum amount of money that the firmcan make by selling its product. We canfind the maximum revenue by setting thederivative of the revenue function equalto zero and solving for x.The derivative of the revenue function isgiven by:R'(x) = 1250 (6/5)xSetting this equal to zero and solving forx, we get:x= 1041.67 ≈ 1042Putting the value of x in revenue finctionwe get,R(x) = $651,041.60Therefore, the maximum revenue is$651,041.60, which occurs when thecompany sells 1042 units.Step 4: Form the required functionThe profit function can be find out by thedifference between the revenue functionand the cost function. It is given by:P(x) = R(x) - C(x)Subbing the revenue and cost functions,we get:32P(x)= 1250x – (²)x² – [28,000 + (²)x²-)x² +1028x-28,000P(x)=To find the maximum profit, we caninvolve similar technique as weaccomplished for tracking down themaximum revenue. Setting the derivativeof the profit function equivalent to zeroand solving for x, we get:2(²)x + 1028 = 0P'(x):= -Solving for x, we get:x = 2570Putting the value of x in profit functionwe get,Max profit = $1,292,980Therefore, the maximum profit is$1,292,80, which occurs when thecompany sells 2570 units.Solutiona) Thus breakeven point = 28 unitb) the maximum revenue is $651,041.60,which occurs when the company sells1042 units.c) the maximum profit is $1,292,80, whichoccurs when the company sells 2570units. +222Suppose a company has fixed costs of $28,000 and variable cost per unit ofdollars, where x is the total number of units produced. Suppose further that the sellingprice 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?

Suppose a company has fixed costs of $28,000 and variable cost per unit of dollars, where x is the…

Answered: d) What price will maximize the profit?

Business

Economics

d) What price will maximize the profit?

ENGR.ECONOMIC ANALYSIS

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ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

Problem 1QTC

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Concept explainers

Profit Maximization

Economic activity is related to the use of scarce resources of the economy that are used for producing goods and services which are required to satisfy various demands of the people. In other words, it can be said that it is all about satisfying the unlimited demands of individuals with the limited resources available. The production decision is taken by the producers or the firms. In economics, the terms ‘producers’, ‘suppliers’, and ‘firms’ are interchangeable. Now the question arises, what does a firm actually mean?

Question

100%

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.

+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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