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
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.
1 answer
1 answer