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# Calculus/Calculus, maximum revenue

Question
a theatre determines that if the admission price is \$10, it averages 100 people in attendance . but for every increase of \$2, it loses 5 customers from the average number.every customer spends an average of \$2 on concessions. what admission price should the theater charge in order to maximize the revenue?

We may express the admission price as 10 + 2x , where x is the number of times the price is raised by \$2.

The number of customers will then be 100 - 5x .

Revenue is then  R(x) = (100 - 5x)(10 + 2x) + (100 -5x)(2) , where (100 -5x)(2) is added because each customer spends \$2 on  concessions.

Simplifying this expression for revenue gives

R(x) = (100 - 5x)(12 + 2x) and

R(x) = (10)(20 - x)(6 + x)

You can use calculus to find the value for x that gives the maximum , or just recognize that the graph of this function is a parabola that opens down and has its highest point when x is the average of the roots .

0 =  (10)(20 - x)(6 + x) gives x = 20 and x = -6 for the roots.

Thus the revenue is a maximum when x =  (20 - 6 ) / 2 = 14/2 = 7

The admission price will be 10 + (2) (7) = 10 + 14 = 24

The maximum revenue occurs when the admission price is \$24

Calculus

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#### Socrates

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I can answer questions from the standard four semester Calculus sequence. I am not prepared for questions on Tensor Calculus. Everything else is welcome. Derivatives, partial derivatives, ordinary differential equations, single and multiple integrals, change of variable, vector integration (Green`s Theorem, Stokes, and Gauss) and applications.

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Ph.D. in Mathematics and many years teaching Calculus at state universities.

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B.S. , M.S. , Ph.D.