A farmer estimates that if he harvests his soybean crop now, he will obtain 1,000 bushels, which he can sell at $3.00 per bushel. However, he estimates that this crop will increase by an additional 1,200 bushels of soybeans for each week he delays harvesting, but the price will drop at a rate of 50

Question

A farmer estimates that if he harvests his soybean crop now, he will obtain 1,000 bushels, which he can sell at [latex]\$ 3.00[/latex] per bushel. However, he estimates that this crop will increase by an additional 1,200 bushels of soybeans for each week he delays harvesting, but the price will drop at a rate of 50 cents per bushel per week; in addition, it is likely that he will experience spoilage of approximately 200 bushels per week for each week he delays harvesting. When should he harvest his crop to obtain the largest net cash return, and how much will be received for his crop at that time? (2.3)

Accepted Answer

Let [latex]X=[/latex] number of weeks to delay harvesting and [latex]R=[/latex] total revenue as a function of [latex]X[/latex]

[latex]\begin{aligned}R &=(1,000 ext { bushels }+1,000 ext { bushels } imes X)(\$ 3.00 / ext { bushel }-\$ 0.50 / ext { bushel } imes X) \R &=\$ 3,000+\$ 2,500 X-\$ 500 X^{2} \ \frac{d R}{d X} &=2,500-1,000 X=0\end{aligned}[/latex]

Solving for [latex]X[/latex] yields [latex]X^{*}=2.5[/latex] weeks

[latex]\frac{\mathrm{d}^{2} R}{\mathrm{~d} X^{2}}=-1,000[/latex] so, we have a stationary point, [latex]X^{*}[/latex], that is a maximum.

Maximum revenue [latex]=\$ 3,000+\$ 2,500(2.5)-500(2.5)^{2}=\$ 6,125[/latex]

Question 2.T-Y-S.B: A farmer estimates that if he harvests his soybean crop now,...

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