
Concept explainers
(a)
To express: the revenue R as a function of x .
(a)

Answer to Problem 39RE
The expression of the revenue R is R=−110x2+150x
Explanation of Solution
Given:
p=−110x+150 0≤x≤1500
Calculation:
By the law of economics, the revenue R is given by R=xp .
Thus,
R=x(−110x+150)=−110x2+150x
Therefore, R=−110x2+150x
Conclusion:
The expression of the revenue R is R=−110x2+150x
(b)
To find:the revenue if 100units are sold
(b)

Answer to Problem 39RE
The revenue if 100 units are sold is $14000
Explanation of Solution
Calculation:
Substitute 100 for x in R=−110x2+150x .
R=−110(100)2+150(100)=−110(1000)+15000=−1000+15000=$14000
Therefore, the revenue if 100 units is sold is $14000 .
Conclusion:The revenue if 100 units is sold is $14000
(c)
To find: the maximum revenue and the quantity x that maximizes the revenue
(c)

Answer to Problem 39RE
When x=750 there will be maximum revenue
The maximum revenue will be $56,250
Explanation of Solution
Calculation:
The function of R is a
So, the revenue will be maximum when the quantity sold x=−b2a .
x=−1502(−110)=15015=750
Therefore, when x=750 there will be maximum revenue.
Find the maximum revenue by substituting 750 for x in R=−110x2+150x .
R=−110(750)2+150(750)=−110(562,500)+112,500=−56,250+11,2500=$56,250
Therefore, the maximum revenue will be $56,250 .
Conclusion:
Therefore, there will be maximum revenuewhen x=750
The maximum revenue will be $56,250
(d)
To find:price charged by the company to maximize revenue?
(d)

Answer to Problem 39RE
$75 should be the price to maximize the revenue
Explanation of Solution
Calculation:
Substitute 750 for x in p=−110x+150 to get the price to maximize the revenue.
p=−110(750)+150=−75+150=$75
Therefore, $75 should be the price to maximize the revenue.
Conclusion:
$75 should be the price to maximize the revenue
Chapter 3 Solutions
Precalculus
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