A 135-kg steer gains 3.5kg/day and costs 80cents/day to keep. The market price for beef cattle is $1.65/kg, but the price falls by 1cent/day. When should the steer be sold to maximize profit.
help is greatly appreciated :)
the answer is susposed to be 52 days, but i can't figure out how to get that.
Oct 12th 2005, 12:07 AM
Profit = Income minus Expenses
Income = (Total weight of steer, in kg) times (unit price, in $/kg)
Let x = number of days from now when the steer is sold -------------------***
The total weight after x days = (135 +3.5*x) kg
The unit price per kg, after x days = $1.65 -$0.01*x = (1.65 -0.01x) dollars
The expenses after x days = $0.80*x = 0.80x dollars
Profit = Income minus expenses
P = (135 +3.5x)(1.65 -0.01x) -0.80x
P = (135*1.65 -135*0.01x +3.5x*1.65 -3.5x*0.01x) -0.80x
P = 222.75 -1.35x +5.775x -0.035x^2 -0.80x
P = 222.75 +3.625x -0.035x^2 ---------------(i)
Now here, we can solve for maximum P by
a) using the properties of a parabola (because Eq.(i) is a vertical parabola that opens downward and so its vertex is its highest point which gives the maximum P.)
or,b) using Calculus. P is maximum or minimum when dP/dx = 0.
Let me assume you know Calculus, so,
Differentiate both sides of (i) with respect to x,
dP/dx = 3.625 -(0.035)[2x]
dP/dx = 3.625 -0.070x
Set that to zero,
0 = 3.3625 -0.07x
0.07x = 3.625
x = 3.625/0.07
x = 51.7875
or, x = 52 days ---------answer.