SOLUTION: A 135-kg steer gains 3.5kg/day and costs 80 cents /day to keep. The market price for beef cattle is $1.65/kg, but the price falls by 1 cent /day. When should the steer be sold to m
Algebra ->
Quadratic Equations and Parabolas
-> SOLUTION: A 135-kg steer gains 3.5kg/day and costs 80 cents /day to keep. The market price for beef cattle is $1.65/kg, but the price falls by 1 cent /day. When should the steer be sold to m
Log On
Question 169103: A 135-kg steer gains 3.5kg/day and costs 80 cents /day to keep. The market price for beef cattle is $1.65/kg, but the price falls by 1 cent /day. When should the steer be sold to maximize profit ?
I tried to make the profit equation for this problem but got confused when I tried to make the cost part of it , how do I make the profit equation? Answer by scott8148(6628) (Show Source):
You can put this solution on YOUR website! the selling price of a steer in x days is the weight (135+3.5x) times the unit price (1.65-.01x)
the cost to keep the steer for x days is .8x
the profit is the selling price minus the cost __ (135+3.5x)(1.65-.01x)-.8x __ this is the function to maximize
FOILing __ 222.75+5.775x-1.35x-.035x^2-.8x = p __ -.035x^2+3.625x+222.75 = p
the maximim is on the axis of symmetry __ the general eqn is x=-b/(2a)
__ in this case __ x=-3.625/[2(-.035)] __ x=51.8 (approx)
the steer should be sold on the 52nd day to maximize profit
