Grade 11 U math
posted by Jamie on .
A 135kg 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= RevenueExpenses.
show the revenue and costs.
put in vertex form.

x=days
135kg +3.5kg/day*x =weight
SalePrice= $(1.650.01*x)/kg*weight
cost= $0.8/day*x
profit=Salepricecost
profit(x)=(1.650.01x)(135 +3.5x)0.8x
profit(x)= 1.65*135+1.65*3.5x0.01*135x0.01*3.5x^20.8x
= 222.750+4.4250x0.035x^20.8x
= 222.750+3.625x0.035x^2
set the derivative to 0 to find max profit
dprofit(x)/dx=0
3.6250.07x=0
x=3.625/0.07
x=51.7857~=52
hope this helps