Question

Farmers can get $8 per bushel for their potatoes on July 1 and after that the price drops by 8 cents per day. On July 1, a farmer has 80 bushels of potatoes in the field and estimates the crop is increasing at a rate of 1 bushel per day. When should the farmer harvest the potatoes to maximize his revenue? (This involves derivatives somewhere)

Answer 1

probably not calculus

Answer 2

call \(x\) the number of days after july 1
the price will be
\[8-.08x\] and the yield will be \(80+x\)

Answer 3

maximize
\[(8-.08x)(80+x)\] which needs no calc as it is a parabola that opens down
find the vertex to get the day you should sell and the amount you make

Answer 4

It is in my calculus book and i already knew how to get those equations i just need help on how to solve it.

Answer 5

nevermind thank you

Answer 6

First, let's express profits as a function of time denoted by x :
\[Profits(x)=(8-0.08x)(80+x)=-0.08x^2+1.6x+640\]This is an upside down parabola and in the context of the problem, we have to find its vertex. Using calculus, we want to find a value of x for which the slope of the parabola is equal to zero. The slope of a function at a given point is the value of its derivative at said point, hence :
\[\frac{ d }{ dx }Profits(x)=-0.16x+1.6\]We solve for x when the derivative is equal to zero :
\[-0.16x+1.6=0 \rightarrow x=10\]Therefore we conclude that the farmer should wait ten days before harvesting.