The theater make the maximum profit at d = $25. Then the maximum profit of the theatre is $ 900.
<h3>What is differentiation?</h3>
The rate of change of a function with respect to the variable is called differentiation. It may be increasing or decreasing.
A community theater uses the function P(d) = (− 4d + 40) (d − 40) to model the profit (in dollars) expected on a weekend when the tickets to a comedy show are priced at d dollars each.
Then the maximum profit of the theatre will be
The function is P(d) = (− 4d + 40) (d − 40)
Differentiate the function with respect to d and put it equal to zero for maximum or minimum profit.

Then the checking for maximum or minimum, again differentiate, we have

The value is less than zero hence maximum value will occur at d = 25.
Then maximum profit will be
P(d) = (− 4×25 + 40) (25 − 40)
P(d) = (− 100 + 40) (−15)
P(d) = (− 60) (− 15)
P(d) = $ 900
More about the differentiation link is given below.
brainly.com/question/24062595
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