A 135 kg steer gains 3.5 kg/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 for max profit?
let x be the number of days before the steer is sold. the money is in units of cents
Profit = Revenue - Cost
we want to maximize Profit. Now,
Cost = 80x
Revenue = 165(135 + 3.5x) - x
now find the formula for the profit and maximize it. do you understand how i got those equations?
Here is one way.
Revenue = (weight)*(cost per unit weight)
If the cow is sold today, R = 135*($1.65)
But the cow is to be sold x days from today.
So the weight ater x days is 135 +3.5*x
cost per unit weight falls by $0.01 per day
So after x days, the price will be (1.65 -0.01x) .....in dollars
Hence the total revenue after x days is
R(x) = (135 +(3.5)x)*(1.65 -0.1x) ...........in dollars
where R(x) is R as a function of x.
The cost per day is $0.80, and it does not depend on how days from now, meaning, it is fixed at $0.80 per day, so, the total cost for x days will be
C(x) = ($0.80)*x
Now, profit = revenue -cost, so,
P(x) = R(x) -C(x)
P(x) = (135 +3.5x)(1.65 -0.01x) -0.80x
P(x) = 222.75 +3.625x -0.035x^2
That is vertical parabola that opens downward---because of the negative x^2.
To the vertex is a maximum.
I assume you don't know Calculus yet, so you can solve for the maximum P by using the properties of a parabola.
I assume also that you know the formula x = -b/(2a) for the x-coordinate of the vertex, so,
x = -(3.625) / 2(-0.035) = 51.786 days
Therefore, for maximum profit, the cow should be sold in 52 days from now. ----answer.