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In any optimization problem involving yields, the first step is to define the yield function. The yield function represents how the yield (output) changes as we alter an input variable. In the cherry tree example, the input is the number of trees per acre, denoted by \(x\). If \(x\) is 30 or fewer, the yield per tree is constant at 50 pounds, and the total yield \(Y\) is \(50x\). This gives us a straightforward linear function.
For more than 30 trees, the yield per tree decreases due to overcrowding. Here, every additional tree over 30 reduces the yield of every tree by 1 pound. Hence, the yield per tree becomes \(50 - (x - 30)\). Therefore, the total yield \(Y\) is expressed as \((50 - (x - 30))x\), which simplifies to \(80x - x^2\). This is a quadratic function that accurately represents the relationship between the number of trees and the total yield.

Derivatives are a core tool in calculus used to analyze functions for slope or rate of change. They are crucial for finding maximum or minimum values in optimization problems. By finding the derivative of the yield function, we can determine how the yield changes with each additional tree.
In our cherry tree problem, we took the derivative of the quadratic yield function \(Y = 80x - x^2\). The derivative \(dY/dx\) helps us find where the slope of the yield function is zero, i.e., where the yield has a maximum or minimum point. Calculating this, we find \(dY/dx = 80 - 2x\). This derivative helps set up the next critical step of identifying critical points.

Critical points occur where the derivative of a function is zero or undefined. These points are potential candidates for local maxima or minima. In optimization problems, especially those involving physical constraints like our cherry yield problem, finding these points is essential.
By setting \(dY/dx = 0\), we solve the equation \(80 - 2x = 0\) to find \(x = 40\). This tells us that planting 40 trees per acre is a critical point for yield. In this context, it suggests an optimal planting strategy to maximize yield.

Maximizing a yield function involves correctly identifying and evaluating critical points. After finding a critical point like \(x = 40\), we substitute it back into the yield function to determine the corresponding yield.
Substituting \(x = 40\) into the equation \(Y = 80x - x^2\) gives us \(Y = 80(40) - 40^2 = 1600\). Thus, by planting 40 trees, the maximum yield per acre is 1600 pounds. This complete process demonstrates not only how to find optimal values but also confirms the logic of the yield function crafted from the beginning. Understanding this helps in similar real-world situations where maximizing output is the goal.