Problem 16 - Calculus for Biology and Medicine

Question

Optimizing Crop Yield A farmer is trying to optimize the amount of nitrogen fertilizer to add to a field. She finds that her yield per square meter increases with the amount, \(N\), of nitrogen added according to the formula: $$ Y(N)=\frac{1}{e^{-N}+1} \quad N \geq 0 $$ (a) Show that \(Y(N)\) increases monotonically with \(N\). If the cost of fertilizer is not important, this result suggests that she should add as much fertilizer as she can to the field. (b) Show either by using calculus, or by making a plot on a graphing calculator, that \(Y(N)\) is concave downward, so there are diminishing returns from using more fertilizer. (c) Suppose that the farmer includes the cost of fertilizer when determining the optimal amount to use. If the cost of one unit of fertilizer is \(C\), then her return, \(N\), becomes: $$ r(N)=Y(N)-C N=\frac{1}{e^{-N}+1}-C N \quad N \geq 0 $$ You may assume that \(C\) is a positive constant. Explain in words why, whatever the value of \(C\) is, we would expect there to be an optimal value of \(N\) that maximizes the return \(r(N)\). (d) Calculate the optimal value of \(N\) if \(C=1 / 8\) (Hint: To find when \(r^{\prime}(N)=0\), make the substitution \(u=e^{-N}\) and solve for \(u .\) You will need to use the quadratic formula.) (e) If \(C=1\) show that the optimal amount of fertilizer for the farmer to add is \(N=0\).

Step-by-Step Solution

Answer

(a) \(Y(N)\) increases monotonically. (b) \(Y(N)\) is concave downward. (d) \(N=-\ln(3-\sqrt{8})\) for \(C=1/8\). (e) Optimal \(N=0\) when \(C=1\).

Step 1: Show Monotonic Increase of Yield Function

To demonstrate that \(Y(N)\) increases monotonically, we compute the derivative \(Y'(N)\) with respect to \(N\). The function is given by \(Y(N) = \frac{1}{e^{-N} + 1}\). Differentiating with respect to \(N\) yields:\[Y'(N) = \frac{d}{dN} \left( \frac{1}{e^{-N} + 1} \right) = \frac{e^{-N}}{(e^{-N} + 1)^2}\]Since \(e^{-N} > 0\) for \(N \geq 0\), it follows that \(Y'(N) > 0\) for all \(N \geq 0\). Thus, \(Y(N)\) is monotonically increasing for \(N \geq 0\).

Step 2: Show Concave Down Nature of Yield Function

To demonstrate concavity, we need to show that the second derivative \(Y''(N)\) is negative. We find the second derivative as follows:\[Y''(N) = \frac{d}{dN} \left(\frac{e^{-N}}{(e^{-N} + 1)^2} \right) = \frac{e^{-N} \cdot (e^{-N} - 1)}{(e^{-N} + 1)^3}\]Since both \(e^{-N} > 0\) and \(e^{-N} - 1 < 0\) for \(N > 0\), \(Y''(N) < 0\), indicating that \(Y(N)\) is concave down, confirming diminishing returns on increasing \(N\).

Step 3: Explain Existence of Optimal Value for Economic Return

The economic return \(r(N)\) is given by \(r(N) = Y(N) - C \cdot N\). Although \(Y(N)\) increases, the term \(-C \cdot N\) causes \(r(N)\) to eventually decrease as \(N\) continues to increase due to the linear cost increase. The diminishing returns of \(Y(N)\) along with the linear increase of costs due to \(N\) ensures that there exists an optimal \(N\) where the derivative \(r'(N)=0\).

Step 4: Calculate Optimal \(N\) for Cost \(C=\frac{1}{8}\)

First, to find \(r'(N)=0\), differentiate \(r(N) = \frac{1}{e^{-N}+1} - \frac{N}{8}\):\[r'(N) = \frac{e^{-N}}{(e^{-N}+1)^2} - \frac{1}{8} = 0\]Substituting \(u=e^{-N}\), gives:\[\frac{u}{(u+1)^2} = \frac{1}{8} \, \Rightarrow \, 8u = (u+1)^2 \, \Rightarrow \, 8u = u^2 + 2u + 1 \, \Rightarrow \, u^2 - 6u + 1 = 0\]Applying the quadratic formula, \(u = \frac{6 \pm \sqrt{36 - 4}}{2} = 3 \pm \sqrt{8}\). Choose \(u=3-\sqrt{8}\) since \(u>0\), and then \(N=-\ln(3-\sqrt{8})\).

Step 5: Show Optimal Amount for \(C=1\) is \(N=0\)

When \(C=1\), differentiate \(r(N) = \frac{1}{e^{-N}+1} - N\):\[r'(N) = \frac{e^{-N}}{(e^{-N}+1)^2} - 1\]Setting \(r'(N) = 0\) implies \(\frac{e^{-N}}{(e^{-N}+1)^2} = 1\). Since it cannot balance, the optimal \(N\) must satisfy decreasing \(r(N)\) for any \(N>0\). Hence, \(N=0\) maximizes \(r(N)\).

Key Concepts

Monotonic Increase

In optimization problems, especially in calculus, understanding whether a function increases or decreases can be crucial. A function is said to be **monotonically increasing** if its value never decreases as the variable increases. In simpler terms, the function keeps climbing or staying the same as you move along the x-axis.
To ascertain if the yield function \( Y(N) = \frac{1}{e^{-N} + 1} \) is monotonically increasing, we examine its derivative. The derivative here, denoted as \( Y'(N) \), is given by:

• \( Y'(N) = \frac{e^{-N}}{(e^{-N} + 1)^2} \)

For \( N \geq 0 \), the expression \( e^{-N} \) is always positive, as exponentiating any negative number results in a positive outcome. Therefore, the derivative of \( Y(N) \) is also positive for all \( N \geq 0 \). This tells us that the function keeps increasing as more fertilizer is added, irrespective of how much is already present. Thus, the yield function is monotonically increasing.

Concave Function

When discussing functions in calculus, the idea of a function being **concave** or **convex** often arises. A concave function bends inwards, much like the inside of a bowl. This property can be identified by examining the second derivative.
For the function \( Y(N) = \frac{1}{e^{-N} + 1} \), we found the second derivative:

• \( Y''(N) = \frac{e^{-N} \cdot (e^{-N} - 1)}{(e^{-N} + 1)^3} \)

Since \( e^{-N} > 0 \) and \( e^{-N} - 1 < 0 \) for any \( N > 0 \), the multiplication results in something negative, which means \( Y''(N) < 0 \). A negative second derivative indicates the function curves downwards, akin to a downhill ski slope. This curvature means each extra increment of \( N \), beyond a certain point, yields progressively smaller increases in output—hence, the concept of diminishing returns. The output keeps growing, but at a slower and slower rate.

Economic Return

The concept of **economic return** is pivotal when assessing farming practices or any investment scenario, blending costs and benefits to find the best outcome. Here, the farmer’s decision-making depends not only on how much crop yield she gets, but also on the costs incurred.
The return function is given by:

• \( r(N) = Y(N) - C \cdot N = \frac{1}{e^{-N} + 1} - C \cdot N \)

This formula represents net profit, obtained by subtracting the cost (\( C \cdot N \)) of applying nitrogen from the yield \( Y(N) \). While more nitrogen increases yield, it also increases costs linearly. As the fertilizer increases indefinitely, the cost can overshadow the benefits due to the diminishing improvement in yield.
Using calculus, specifically by finding when the derivative \( r'(N) = 0 \), it's possible to determine the optimal amount of fertilizer, considering cost constraints. For example, when \( C = 1/8 \), there’s a clear optimal point beyond which adding more does not maximize return. Similarly, if \( C = 1 \), using maximum calculus techniques indicates that no fertilizer results in the highest profit, as costs would rise too sharply otherwise. This balance between costs and returns underpins many practical economic decisions.