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A company operates 16 oil wells in a designated area. Each pump on average, extracts 240 barrels of oil daily. The company can add more wells but every added well reduces the average daily output of each of the wells by 8 barrels. How many wells should the company add in order to maximize daily production?
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Calculus 1 / AB
Calculus 2 / BC
Applications of Differentiation
Missouri State University
Idaho State University
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were asked to answer a question about drilling oil wells using optimization problems. Mhm, it said. Oh, okay, So we're told that a company operates 16 oil wells in a designated area. Each pump on average extracts 240 barrels of oil daily, and the company can add more wells. But every added well reduces the average daily output of each of the wells by eight barrels were asked How many wells should the company ad in order to max maximize daily productivity? Okay, if she so an expression for the total number of Wells T is going to be the initial number of wells 16 plus the additional number of wells X, an expression for the number of barrels. And this is the 240 initial barrels minus eight x so minus eight for each well added an expression for the total output. Oh, this is uh huh. The number of wells T. Spencer times we number of barrels. What's up? Per pump, which is n mhm. So this is 16 plus x times 240 minus eight x. I got this. No, the entire sound. This is negative. Eight X squared. It's a word that wasn't missed. Plus 112 x plus 3840. Now, in order to maximize our daily production, we're going to find the derivative Oh, prime. As a function of X, this is negative 16 x plus 112. You want to set this equal to zero, and so we get X is equal to seven. Now, we also find o double prime of X second derivative. This is negative 16, which is always less than zero. This implies that it is oh, has a maximum at X equals seven by the second derivative test. And therefore, our answer is simply add seven wells, Yeah.
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