Hercules Films is deciding on the price of the video release of its film Bride of the Son of Frankenstein. Marketing estimates that at a price of p dollars, it can sell g = 300,000 – 15,000p copies, but each copy costs $4 to make. (a) Find the cost, revenue and profit functions. C(p) = 1200000 – 60000p R(p)= 300000p – 15000p P(p)= -15000p + 360000p – 1200000 (b) What price will give the greatest profit? p = 12 dollars Second derivative test: Your answer above is a critical point for the profit function. To show it is a maximum, calculate the second derivative of the profit function. P"(p)= -1 Evaluate P"(p) at your critical point. The result is negative which means that the profit is concave down v at the critical point, and the critical point is a maximum.

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
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Chapter1: Expressions And Functions
Section1.8: Interpreting Graphs Of Functions
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Hercules Films is deciding on the price of the video release of its film Bride of the Son of Frankenstein. Marketing estimates that at a price of p dollars, it
can sell g = 300,000 – 15,000p copies, but each copy costs $4 to make.
(a) Find the cost, revenue and profit functions.
C(p) = 1200000 – 60000p
R(p)= 300000p – 15000p
P(p)= -15000p + 360000p – 1200000
(b) What price will give the greatest profit?
p = 12
dollars
Second derivative test:
Your answer above is a critical point for the profit function. To show it is a maximum, calculate the second derivative of the profit function.
P"(p)= -1
, which means that the profit is concave down v
Evaluate P"(p) at your critical point. The result is negative
critical point is a maximum.
at the critical point, and the
Transcribed Image Text:Hercules Films is deciding on the price of the video release of its film Bride of the Son of Frankenstein. Marketing estimates that at a price of p dollars, it can sell g = 300,000 – 15,000p copies, but each copy costs $4 to make. (a) Find the cost, revenue and profit functions. C(p) = 1200000 – 60000p R(p)= 300000p – 15000p P(p)= -15000p + 360000p – 1200000 (b) What price will give the greatest profit? p = 12 dollars Second derivative test: Your answer above is a critical point for the profit function. To show it is a maximum, calculate the second derivative of the profit function. P"(p)= -1 , which means that the profit is concave down v Evaluate P"(p) at your critical point. The result is negative critical point is a maximum. at the critical point, and the
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