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The cost of controlling emissions at a firm rises rapidly as the amount of emissions reduced increases. Here is a possible model: The daily cost in dollars to reduce emissions by q pounds of pollutant in a day is given by C(q) = 4,400 + 100q?. (a) What is the average daily cost per pound when emissions are reduced by g pounds in a day? C(9) = (b) What level of reduction corresponds to the lowest average daily cost per pound of pollutant? (Round your answer to two decimal places.) pounds of pollutant What would be the resulting minimum average daily cost per pound? (round to the nearest dollar) dollars Second derivative test: Your answer above is a critical point for the average daily cost function. To show it is a minimum, calculate the second derivative of the average daily cost function. Evaluate C"(q) at your critical point. The result is positive , which means that the average cost is concave up at the critical point, and the critical point is a minimum.

The cost of controlling emissions at a firm rises rapidly as the amount of emissions reduced increases. Here is a possible model: The daily cost in dollars to reduce emissions by q pounds of pollutant in a day is given by C(q) = 4,400 + 100q?. (a) What is the average daily cost per pound when emissions are reduced by g pounds in a day? C(9) = (b) What level of reduction corresponds to the lowest average daily cost per pound of pollutant? (Round your answer to two decimal places.) pounds of pollutant What would be the resulting minimum average daily cost per pound? (round to the nearest dollar) dollars Second derivative test: Your answer above is a critical point for the average daily cost function. To show it is a minimum, calculate the second derivative of the average daily cost function. Evaluate C"(q) at your critical point. The result is positive , which means that the average cost is concave up at the critical point, and the critical point is a minimum.

College Algebra
College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter1: Equations And Graphs
Section1.3: Lines
Problem 92E
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The cost of controlling emissions at a firm rises rapidly as the amount of emissions reduced increases. Here is a possible model: The daily cost in dollars to reduce emissions by q pounds of pollutant in a day is given by C(q) = 4,400 + 100g?. (a) What is the average daily cost per pound when emissions are reduced by g pounds in a day? C(q) = (b) What level of reduction corresponds to the lowest average daily cost per pound of pollutant? (Round your answer to two decimal places.) pounds of pollutant What would be the resulting minimum average daily cost per pound? (round to the nearest dollar) dollars Second derivative test: Your answer above is a critical point for the average daily cost function. To show it is a minimum, calculate the second derivative of the average daily cost function. C"(q)= Evaluate C"(q) at your critical point. The result is positive , which means that the average cost is concave up at the critical point, and the critical point is a minimum.
Transcribed Image Text:The cost of controlling emissions at a firm rises rapidly as the amount of emissions reduced increases. Here is a possible model: The daily cost in dollars to reduce emissions by q pounds of pollutant in a day is given by C(q) = 4,400 + 100g?. (a) What is the average daily cost per pound when emissions are reduced by g pounds in a day? C(q) = (b) What level of reduction corresponds to the lowest average daily cost per pound of pollutant? (Round your answer to two decimal places.) pounds of pollutant What would be the resulting minimum average daily cost per pound? (round to the nearest dollar) dollars Second derivative test: Your answer above is a critical point for the average daily cost function. To show it is a minimum, calculate the second derivative of the average daily cost function. C"(q)= Evaluate C"(q) at your critical point. The result is positive , which means that the average cost is concave up at the critical point, and the critical point is a minimum.
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