Question 8.3.1: Suppose Y(N) bushels of wheat are harvested per acre of land......
Suppose Y(N) bushels of wheat are harvested per acre of land when N pounds of fertilizer per acre are used. If P is the dollar price per bushel of wheat and q is the dollar price per pound of fertilizer, then profits in dollars per acre are
π(N) = PY(N) − qN
for N ≥ 0. Suppose there exists N^{∗} such that π^{\prime}(N) ≥ 0 for N ≤ N^{∗}, whereas π^{\prime}(N) ≤ 0 for N ≥ N^{∗}. Then N^{∗} maximizes profits, and π^{\prime}(N^{∗}) = 0. That is, PY^{\prime}(N^{∗}) − q = 0, so
PY^{\prime}(N^{∗}) = q (∗)
Let us give an economic interpretation of this condition. Suppose N^{∗} units of fertilizer are used and we contemplate increasing N^{∗} by one unit. What do we gain? If N^{∗} increases by one unit, then Y(N^{∗} + 1) − Y(N^{∗}) more bushels are produced. Now Y(N^{∗} + 1) − Y(N^{∗}) ≈Y(N^{∗}). For each of these bushels, we get P dollars, so by increasing N^∗ by one unit, we gain approximately PY^{\prime}(N^{∗}) dollars. On the other hand, by increasing (N^{∗}) by one unit, we lose q dollars, because this is the cost of one unit of fertilizer. Hence, we can interpret (∗) as follows: In order to maximize profits, you should increase the amount of fertilizer to the level (N^{∗}) at which an additional pound of fertilizer equates the changes in your gains and losses from the extra pound.
(a) In an (unrealistic) example, suppose that Y(N) =\sqrt{N}, P = 10, and q = 0.5. Find the amount of fertilizer which maximizes profits in this case.
(b) An agricultural study in Iowa estimated the yield function Y(N) for the year 1952 as
Y(N) = −13.62 + 0.984N − 0.05N^{1.5}
If the price of wheat is $1.40 per bushel and the price of fertilizer is $0.18 per pound, find the amount of fertilizer that maximizes profits.
(a) The profit function is
\pi(N)=P Y(N)-q N=10N^{1/2}-0.5N,\;N\geq0Then π^{\prime}(N) = 5N^{−1/2} − 0.5. We see that π^{\prime}(N^{∗}) = 0 when (N^{∗})^{−1/2} = 0.1, hence N^{∗} = 100. Moreover, it follows that π^{\prime}(N) ≥ 0 when N ≤ 100 and π^{\prime}(N) ≤ 0 when N ≥ 100. We conclude that N^{∗} = 100 maximizes profits. See Fig. 8.3.1.
(b) In this case,
\pi(N)=1.4(-13.62+0.984N-0.05N^{1.5})-0.18N \\ \quad\quad=-19.068+1.1976N-0.07N^{1.5}so that
\pi^{\prime}(N)=1.1976-0.07\cdot1.5N^{0.5}=1.1976-0.105\sqrt{N}{\mathrm{Hence}}\;\pi^{\prime}(N^{*})=0{\mathrm{~when~}}0.105{\sqrt{N^{*}}}=1.1976. This implies that
\sqrt{N^{*}}=1.1976/0.105\approx11.4\ \mathrm{and~so}\ \ N^{*}\approx(11.4)^{2}\approx130By studying the expression for π^{\prime}(N), we see that π^{\prime}(N) is positive to the left of N^{∗} and negative to the right of N^{∗}. Hence, N^{∗} ≈ 130 maximizes profits. The graph of π(N) is shown in Fig. 8.3.2.
