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The weekly revenue from selling movie tickets at a local theater can be modeled by:

R(p) = 122p*e⁻¹ᴾ hundred dollars, when the ticket price is p dollars/ticket. 3 ≤ p ≤ 12.
Determine the absolute maximum and absolute minimum revenue. You must show the work that confirms these are the extrema.

User Tom Bates
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Final answer:

The absolute maximum and minimum revenue can be determined by finding the critical points of the revenue function and evaluating them within the given interval. The critical points are found by setting the derivative of the revenue function equal to zero. The absolute maximum revenue is approximately $44.04 at p = 1, and the absolute minimum revenue is approximately $0.018 at p = 12.

Step-by-step explanation:

To determine the absolute maximum and minimum revenue, we need to find the critical points of the revenue function and evaluate them within the given interval. To find the critical points, we need to find where the derivative of the revenue function is equal to zero or undefined. The derivative of R(p) = 122p * e^(-p) is R'(p) = 122e^(-p) - 122pe^(-p).

To find the critical points, we set R'(p) equal to zero and solve for p: 122e^(-p) - 122pe^(-p) = 0. Factoring out the common factor e^(-p), we get 122e^(-p)(1 - p) = 0. This equation is satisfied when e^(-p) = 0 or when (1 - p) = 0. Since e^(-p) is never zero for real values of p, the only solution is (1 - p) = 0. Solving for p, we find p = 1.

Now that we have the critical point p = 1, we can evaluate the revenue at the critical point as well as the endpoints of the given interval.

  • R(3) = 122(3)e^(-3) = 122(3)e^(-3) ≈ 29.43
  • R(12) = 122(12)e^(-12) = 122(12)e^(-12) ≈ 0.018
  • R(1) = 122(1)e^(-1) = 122(1)e^(-1) ≈ 44.04

Therefore, the absolute maximum revenue is approximately $44.04 at p = 1, and the absolute minimum revenue is approximately $0.018 at p = 12.

User Getitstarted
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