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) = 4,100 + 95q2. (a) What is the average daily cost per pound when emissions are reduced by q 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 pollutantWhat would be the resulting minimum average daily cost per pound? (round to the nearest dollar) _________dollarsSecond 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______________ ( negative or positive), which means that the average cost is ______________ (concave down or concave up) at the critical point, and the critical point is a minimum.
Equations and Inequations
Equations and inequalities describe the relationship between two mathematical expressions.
Linear Functions
A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.
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
(a) What is the average daily cost per pound when emissions are reduced by q pounds in a day?
(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
at your critical point. The result is______________ ( negative or positive), which means that the average cost is ______________ (concave down or concave up) at the critical point, and the critical point is a minimum.
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