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How should ticket prices be set to maximize revenue? (Round your answer to the nearest cent.) The equation is p(x)=(58000-x)/(2000).
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Is that complete?
does this something have anything to do with optimization or calculus?
it's a calculus problem. here is the original. A baseball team plays in a stadium that holds 54,000 spectators. With ticket prices at $10, the average attendance had been 38,000. When ticket prices were lowered to $8, the average attendance rose to 42,000. I'm just asking about part b
Maximum revenue occurs where dp/dx = 0
so i've found p(x). at other places, people said that to find Max revenue, you use r(x)=x*p(x) but i dont know what to do after that :/
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