Problem 27 - Calculus for Biology and Medicine

Question

Suppose crop yield \(Y\) depends on nitrogen \((\mathrm{N})\) and phosphorus (P) concentrations as $$ Y(N, P)=N P e^{-(N+P)} $$ Find the value of \((N, P)\) that maximizes crop yield.

Step-by-Step Solution

Answer

The maximum yield occurs at \((N, P) = (1, 1)\).

Step 1: Understanding the Problem

We are asked to find the point \((N, P)\) that maximizes the function \(Y(N, P) = NP e^{-(N+P)}\). This is a problem of optimization where we need to find the critical points and determine which among these provides a maximum value for the yield.

Step 2: Calculate the Partial Derivatives

We need to find the partial derivatives of \(Y(N, P)\) with respect to \(N\) and \(P\). First, compute \(\frac{\partial Y}{\partial N} \) and \(\frac{\partial Y}{\partial P} \) using the product and chain rule.\[\frac{\partial Y}{\partial N} = (P - NP)e^{-(N+P)}\] \[\frac{\partial Y}{\partial P} = (N - NP)e^{-(N+P)}\]

Step 3: Setting Partial Derivatives to Zero

Set both \(\frac{\partial Y}{\partial N} = 0\) and \(\frac{\partial Y}{\partial P} = 0\) and solve for \(N\) and \(P\). This gives us the system:\[(P - NP)e^{-(N+P)} = 0\] \[(N - NP)e^{-(N+P)} = 0\]Since \(e^{-(N+P)}\) is never zero, we solve:\[P(1-N) = 0\]\[N(1-P) = 0\]

Step 4: Solving the System of Equations

From the system:- \(P(1-N) = 0\) suggests \(P = 0\) or \(N=1\).- \(N(1-P) = 0\) suggests \(N = 0\) or \(P=1\).Therefore, potential critical points are \((N, P) = (1,1), (0, P), (N, 0)\). Evaluate \( (1,1)\).

Step 5: Evaluating and Checking the Critical Point

Substitute into the yield function to verify:\[Y(1, 1) = 1 \times 1 \times e^{-2} = e^{-2} \]Other critical points such as \((N, 0)\) and \((0, P)\) yield zero because one of the factors is zero in each case.

Step 6: Conclusion

Since \(Y(1, 1) = e^{-2}\) is the only feasible critical point that doesn't yield zero, it is most likely the maximum yield. Double check the reasoning with inequality testing or more derivatives if needed for certainty, but \((1, 1)\) is a logical conclusion from the steps.

Key Concepts

Partial Derivatives

When dealing with functions of multiple variables, such as our crop yield function **Y(N, P)**, partial derivatives are an essential tool in calculus. They help us understand how the function changes with respect to one variable while keeping the other constant. In simpler terms, they allow us to "peek" at the function's slope in each direction separately.

For our function, we calculated the partial derivatives with respect to nitrogen **(N)** and phosphorus **(P)**:

• The partial derivative with respect to **N** is \(\frac{\partial Y}{\partial N} = (P - NP)e^{-(N+P)} \),
• The partial derivative with respect to **P** is \(\frac{\partial Y}{\partial P} = (N - NP)e^{-(N+P)} \).

These equations help quantify how a tiny change in either nitrogen or phosphorus affects the overall yield. Computing these derivatives is achieved using common calculus rules for products and exponentials, known as the product rule and the chain rule. These skills are foundational as they provide insights necessary for finding critical points later.

Critical Points

Finding critical points is a crucial aspect of optimization problems in multivariable calculus. A critical point occurs where the partial derivatives of a function are zero or undefined. In our example, to find these points, we set the partial derivatives of our yield function to zero:

• \( (P - NP)e^{-(N+P)} = 0 \)
• \( (N - NP)e^{-(N+P)} = 0 \)

Since the exponential term is never zero, simplifying results in two simpler equations: \( P(1-N) = 0 \) and \( N(1-P) = 0 \).

By solving these, we discover the potential critical points: \( (1,1), (0, P), (N, 0) \). Critical points tell us where to evaluate the function to determine potential maximum or minimum values. They are valuable because they pinpoint where changes zero out, indicating possible extreme values or flatness in the function.

Multivariable Calculus

Multivariable calculus extends the principles of single-variable calculus to functions with two or more independent variables, like our crop yield based on nitrogen and phosphorus concentrations. This branch of calculus allows us to analyze and optimize situations involving multiple changing factors, which is common in real-world problems.

Key techniques in multivariable calculus include:

• Understanding **partial derivatives**, which give the rate of change along each independent variable.
• Locating **critical points** through setting partial derivatives to zero or undefined, providing candidate points for optimization analysis.

The yield function \( Y(N, P) = NP e^{-(N+P)} \) is an example of how these principles apply. As we analyzed its critical points, our ultimate goal was to find the combination of nitrogen and phosphorus that maximizes crop yield.

Multivariable optimization techniques enable this by providing tools and methodologies to handle complex surfaces shaped by multiple variables. This understanding ultimately helps make informed decisions, whether in crop management or other multifactor situations.