Homework Help: Grade 11 U math

Posted by Jamie on Saturday, November 5, 2011 at 9:49pm.

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?

Revenue= (Price/units)(#of units)
Cost=(cost/unit)(# of units) + set costs
Profit= Revenue-Expenses.
show the revenue and costs.
put in vertex form.

Grade 11 U math - JandBsDad, Sunday, November 6, 2011 at 9:47am
x=days

135kg +3.5kg/day*x =weight

SalePrice= $(1.65-0.01*x)/kg*weight

cost= $0.8/day*x

profit=Saleprice-cost

profit(x)=(1.65-0.01x)(135 +3.5x)-0.8x

profit(x)= 1.65*135+1.65*3.5x-0.01*135x-0.01*3.5x^2-0.8x

= 222.750+4.4250x-0.035x^2-0.8x

= 222.750+3.625x-0.035x^2

set the derivative to 0 to find max profit

dprofit(x)/dx=0

3.625-0.07x=0

x=3.625/0.07

x=51.7857~=52

hope this helps

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