Asked by Max
In a large bat factory, marginal revenues increase at first before decreasing rapidly as more bats are made. A limitation to keep in mind is the maximum capacity of the factory being 50 thousand bats per month. If the marginal revenue of making x thousand bats a month is 100-((18x)/((1+0.03x^2)^2)), find the revenue for making 10 thousand bats in a month. How do I enter this answer in in millions of dollars?
Answers
Answered by
Jake
I understand that you find the antiderivative of the marginal revenue function but how do I calculate for C in order to get a function I can plug values into? Thats the part I get confused on.
Answered by
oobleck
You don't need to find C, since the total revenue is a definite integral.
∫[0,10] 100-((18x)/((1+0.03x^2)^2)) dx
= 100x + 30000/(3x^2+1000) [0,10] = 775
∫[0,10] 100-((18x)/((1+0.03x^2)^2)) dx
= 100x + 30000/(3x^2+1000) [0,10] = 775
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