SOLUTION: Abby Normal has a pig that presently weighs 200 pounds. She could sell it now for a price of $1.40 per pound. The pig is gaining 5 pounds per week while the price per pound is dr
Question 1158540: Abby Normal has a pig that presently weighs 200 pounds. She could sell it now for a price of $1.40 per pound. The pig is gaining 5 pounds per week while the price per pound is dropping 2 cents per week. When should Abby sell the pig to get the maximum amount of money for it? What is the maximum profit?
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Abby Normal has a pig that presently weighs 200 pounds. She could sell it now for a price of $1.40 per pound.
The pig is gaining 5 pounds per week while the price per pound is dropping 2 cents per week. When should Abby
sell the pig to get the maximum amount of money for it? What is the maximum profit?
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The weight as a function of time is
W(t) = 200 + 5t pounds
where t is the time in weeks.
The price per pound as a function of time is
P(t) = 1.40 - 0.02t,
The revenue is
W*P = (200+5t)*(1.40-0.02t) dollars (1)
It is a quadratic function with negative coefficient at the quadratic tem, so the quadratic function has a MAXIMUM.
The function has zeroes at t= = -40 and t= = 70.
The maximum of the function is located exactly half-way beteen the zeroes.
So, the maximum is achieved at t= = = 15 weeks.
At this value of t, the quadratic function value is (200+5*15)*(1.40-0.02*15) = 275*1.10 = 302.50 dollars.
ANSWER. To get the maximum amount of money, Abby should sell the pig in 15 weeks.
The maximum amount of money will be 302.50 dollars.
Solved.
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Had Abby sold the pig "today", she would get only 200*1.40 = 280 dollars.
Compare it with the found optimal value of 302.50 dollars (!)
