Asked by anonymous
The monthly revenue R (in thousands of dollars) from the sales of a digital picture frame is approximated by R(p) = −10p^2 + 1480p, where p is the price per unit (in dollars).
a) Find the unit price that will yield a maximum monthly revenue.
answer: -b/2a=-(1480/2(-10))=7400
is this answer correct??
b) What is the maximum monthly revenue?
a) Find the unit price that will yield a maximum monthly revenue.
answer: -b/2a=-(1480/2(-10))=7400
is this answer correct??
b) What is the maximum monthly revenue?
Answers
Answered by
oobleck
1480/20 is greater than 1480? Typo alert!
max value is c - b^2/4a
max value is c - b^2/4a
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