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anonymous
 5 years ago
A manufacturer can produce x units per week. The cost to produce these items is given by y=50x+20,000. These items will set at a unit price given by p=2000.01x. How many items should the manufacturer produce in order to maximize profit? (Remember: revenue = number of units x unit price, and profit = revenue  total cost) (Related Rates)
anonymous
 5 years ago
A manufacturer can produce x units per week. The cost to produce these items is given by y=50x+20,000. These items will set at a unit price given by p=2000.01x. How many items should the manufacturer produce in order to maximize profit? (Remember: revenue = number of units x unit price, and profit = revenue  total cost) (Related Rates)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, the basic approach is to combine the functions you're given into a single profit equation, then find where the derivative = 0 to find a maximum (I'm assuming this is for a calculus class?) So profit = revenue  cost = x(200  0.01x)  50x + 20000 = 200x  0.01x^2  50x + 20000 = 0.01x^2 + 150x + 20000 The first derivative of that is: d/dx (profit) = .02x +50 and that will equal zero when x = 2500

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, woops...the derivative is wrong the constant should be 150, which will make the final answer 7500, I believe

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0given x stands for number of units p stands for unit price y stands for total cost so profit = number of units x unit price  total cost profit = xp  y profit = x(2000.01x)  (50x+20000) now just expand and collect like terms, you should end up with a simplier quadratic equation. To find the max profit, it depends on what you are learning. I believe the most advanced way is using the derivative. Another way is to find the vertex of the quadratic.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right the constant is 150, thanks a lot, i understand it now
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