Lesson When is the best time to sell a pig ?

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When is the best time to sell a pig ?

Problem 1

Abby 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?

Solution

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 money amount to get 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.


Ar this value of t, the quadratic function value is   = 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.

My other lessons in this site on finding the maximum/minimum of a quadratic function are
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    - Briefly on finding the minimum/maximum of a quadratic function
    - HOW TO complete the square to find the vertex of a parabola
    - Briefly on finding the vertex of a parabola
    - A rectangle with a given perimeter which has the maximal area is a square
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    - Using quadratic functions to solve problems on maximizing revenue/profit
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    - Minimal distance between sailing ships in a sea
    - Advanced lesson on finding minima of (x+1)(x+2)(x+3)(x+4)
    - OVERVIEW of lessons on finding the maximum/minimum of a quadratic function

Use this file/link ALGEBRA-I - YOUR ONLINE TEXTBOOK to navigate over all topics and lessons of the online textbook ALGEBRA-I.

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