Problem 27 - Calculus for Biology and Medicine

Question

Suppose crop yield \(Y\) depends on nitrogen \((N)\) 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 crop yield is maximized at \((N, P) = (1, 1)\).

Step 1: Find the Partial Derivatives

To find the maximum crop yield, we first need to calculate the partial derivatives of the function with respect to both nitrogen \((N)\) and phosphorus \((P)\). The function given is \[ Y(N, P) = NPe^{-(N+P)}. \]The partial derivative with respect to \(N\) is:\[ \frac{\partial Y}{\partial N} = Pe^{-(N+P)} - NPe^{-(N+P)}. \]Similarly, the partial derivative with respect to \(P\) is:\[ \frac{\partial Y}{\partial P} = Ne^{-(N+P)} - NPe^{-(N+P)}. \]

Step 2: Set Partial Derivatives to Zero

To find the critical points, set the partial derivatives equal to zero:\[ \frac{\partial Y}{\partial N} = Pe^{-(N+P)} - NPe^{-(N+P)} = 0, \]which simplifies to:\[ P - NP = 0 \rightarrow P(1-N) = 0. \]Similarly, for phosphorus:\[ \frac{\partial Y}{\partial P} = Ne^{-(N+P)} - NPe^{-(N+P)} = 0, \]\[ N - NP = 0 \rightarrow N(1-P) = 0. \]

Step 3: Solve the System of Equations

The equations are:1. \(P(1-N) = 0\)2. \(N(1-P) = 0\)From the first equation, either \(P = 0\) or \(N = 1\). From the second equation, either \(N = 0\) or \(P = 1\).These give possible critical points: \((N, P) = (1, 1), (0, 1), (1, 0), (0, 0)\).

Step 4: Evaluate the Function at Critical Points

Evaluate \(Y(N, P)\) at each of the critical points:- For \((1, 1)\): \[ Y(1,1) = 1 \cdot 1 \cdot e^{-2} = e^{-2}. \]- For \((0, 1)\) and \((1, 0)\): \[ Y(0,1) = 0 \] and \[ Y(1,0) = 0. \]- For \((0, 0)\): \[ Y(0,0) = 0. \]

Step 5: Determine the Maximum Yield

Compare the values obtained from the critical points:\[ Y(1, 1) = e^{-2} \] is the highest value compared to \(Y(0, 1) = 0\), \(Y(1, 0) = 0\), and \(Y(0, 0) = 0\).Thus, the maximum yield occurs at \((N, P) = (1, 1)\) with yield \(e^{-2}.\)

Key Concepts

Partial Derivatives

Understanding what partial derivatives are can unlock numerous optimization problems, like maximizing crop yield. A partial derivative measures how a multivariable function changes as one of the variables changes while keeping the other variables constant. In the context of crop yield, the function is dependent on both nitrogen and phosphorus, so we aim to find how the yield changes when only one of these nutrient levels is varied. When you take the partial derivative of a function like \( Y(N, P) = NPe^{-(N+P)} \) with respect to nitrogen (\(N\)), you treat phosphorus (\(P\)) as a constant, and vice versa. This will lead to two separate equations that can be solved to find places where the rate of change in the crop yield function is zero for both nutrients. That brings us to the critical points.

Critical Points

In optimization, critical points are where the function's derivatives equal zero. These points can possibly indicate maximum, minimum or saddle points in the context of our crop yield function. Once we have the partial derivatives: \[ \frac{\partial Y}{\partial N} = Pe^{-(N+P)} - NPe^{-(N+P)} \] and \[ \frac{\partial Y}{\partial P} = Ne^{-(N+P)} - NPe^{-(N+P)} \], we set them to zero to identify critical points. Solving these equations helps detect where changes in nitrogen or phosphorus no longer affect yield. In this exercise, solving yields critical points such as \((N, P) = (1, 1), (0, 1), (1, 0), (0, 0)\). It's at these points the crop yield stabilizes momentarily, allowing us to check where it might be maximum. Once these points are identified, you would usually test them or apply second derivative tests to classify which of those are maxima.

Crop Yield Maximization

Maximizing crop yields is a crucial concern in agriculture, making it essential to understand how to achieve peak productivity. After calculating the partial derivatives and determining the critical points, the next step involves evaluating the function at these points. For our function, evaluating \( Y(N, P) \) at the critical points: \((1, 1), (0, 1), (1, 0), (0, 0)\), indicates that \((1, 1)\) gives the largest crop yield. So, maximum yield is secured at this point, yielding \( e^{-2} \). Even though zero values are found for other points, knowing how to interpret and evaluate each critical point effectively can make significant strides in agricultural planning. Through evaluating these points, farmers can make evidence-based decisions on nutrient application for optimal crop growth.