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.
b) What is the maximum monthly revenue?
@oobleck you said this:
1480/20 is greater than 1480? Typo alert!
max value is c - b^2/4a
but there is no c, the only a and b
there is always a c. It's just zero!
To find the unit price that will yield a maximum monthly revenue, we can use calculus. The revenue function is given as R(p) = -10p^2 + 1480p.
a) To find the unit price that yields a maximum revenue, we need to find the critical points of the function. Critical points occur when the derivative of the function is equal to zero or undefined. In this case, let's find the derivative of R(p).
R'(p) = d/dp(-10p^2 + 1480p)
= -20p + 1480
Setting R'(p) equal to zero and solving for p:
-20p + 1480 = 0
-20p = -1480
p = 1480/20
p = 74
So, the unit price that will yield a maximum monthly revenue is $74.
b) To find the maximum monthly revenue, we can substitute the value of p = 74 into the revenue function R(p):
R(74) = -10(74)^2 + 1480(74)
= -10(5476) + 109520
= -54760 + 109520
= 54760
Therefore, the maximum monthly revenue is $54,760.