/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 Parasite Propagation Boll weevil... [FREE SOLUTION] | 91Ó°ÊÓ

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Parasite Propagation Boll weevils have long presented a threat to cotton crops in the southern United States. Research has been done to determine the optimal conditions for reproduction of Catolaccus grandis, a parasite that attaches to boll weevils and kills them. The number of eggs produced in 30 days by one female \(C .\) grandis under conditions of 16 hours of light each day, a constant temperature of \(y^{\circ} \mathrm{C},\) and a relative humidity of \(x \%\) can be modeled by $$ \begin{aligned} f(x, y)=&-4191.6877+299.7038 x+23.1412 y \\ &-5.2210 x^{2}-0.0937 y^{2}-0.4023 x y \text { eggs } \end{aligned} $$ (Source: J. A. Morales-Ramos, S. M. Greenberg, and E. G. King, "Selection of Optimal Physical Conditions for Mass Propagation of Catolaccus grandis," Environmental Entomology, vol. \(25,\) no. 1 \((1996),\) pp. \(165-173)\) a. Calculate the point where the partial derivatives of \(f\) are both equal to zero. b. Write a sentence interpreting the point in part \(a\) and characterizing it as a maximum, a minimum, or a saddle point.

Short Answer

Expert verified
The critical point, approximately \((x, y) = (28.74, 61.76)\), is a local maximum.

Step by step solution

01

Find Partial Derivatives

First, we find the partial derivatives of the function \(f(x, y)\) with respect to \(x\) and \(y\). The given function is: \[f(x, y) = -4191.6877 + 299.7038x + 23.1412y - 5.2210x^2 - 0.0937y^2 - 0.4023xy\] The partial derivatives are:- \(\frac{\partial f}{\partial x} = 299.7038 - 10.4420x - 0.4023y\)- \(\frac{\partial f}{\partial y} = 23.1412 - 0.1874y - 0.4023x\)
02

Set Partial Derivatives to Zero

Now, we set both partial derivatives equal to zero to find critical points:\[299.7038 - 10.4420x - 0.4023y = 0 \ 23.1412 - 0.1874y - 0.4023x = 0\]This results in a system of equations to solve for \(x\) and \(y\).
03

Solve System of Equations

Solve the system of equations:1. From the first equation, solve for \(y\): \(y = \frac{299.7038 - 10.4420x}{0.4023}\).2. Substitute \(y\) into the second equation and solve for \(x\).3. Substitute the value of \(x\) back into the equation for \(y\) to find \(y\).Solving these gives approximately \(x = 28.74\) and \(y = 61.76\).
04

Determine the Nature of the Critical Point

To determine whether the critical point is a maximum, minimum, or saddle point, check the second derivatives and use the second derivative test.Calculate the second partial derivatives:- \(\frac{\partial^2 f}{\partial x^2} = -10.4420\)- \(\frac{\partial^2 f}{\partial y^2} = -0.1874\)- \(\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} = -0.4023\)Find the determinant of the Hessian matrix:\[D = \left|\begin{array}{cc} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \ \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2} \end{array}\right| = (-10.4420)(-0.1874) - (-0.4023)^2\]Calculate \(D\) to verify the nature of the point.
05

Conclusion: Interpret the Critical Point

Finish the Hessian determinant calculation. Since \(D > 0\) and \(\frac{\partial^2 f}{\partial x^2} < 0\), the critical point at \((x, y) = (28.74, 61.76)\) is a local maximum. This means these conditions yield the highest expected egg production for Catolaccus grandis in the specified ranges.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Critical Points
Critical points are the values for which the partial derivatives of a function are equal to zero. They are vital in determining areas of interest on a surface defined by a mathematical function, like finding peaks, valleys, or flat areas on a terrain map. In our context, these points help us identify the optimal temperature and humidity levels for the reproduction of Catolaccus grandis, a parasite that targets boll weevils.

After finding the partial derivatives of the function related to the egg production of Catolaccus grandis, we set these derivatives equal to zero. This helps in locating the critical points. For the given function, solving the equations \ \( \frac{\partial f}{\partial x} = 0 \) and \ \( \frac{\partial f}{\partial y} = 0 \) enables us to find the precise combination of temperature and humidity, marked as \ \((x, y) = (28.74, 61.76)\), which could potentially maximize egg production."

It's crucial to understand the concept of critical points because they guide us towards finding the most efficient conditions for biological or physical processes, such as maximizing the propagation of beneficial organisms like Catolaccus grandis.
Hessian Matrix
To further analyze the nature of critical points, we use the Hessian matrix. This matrix is a square matrix of second-order partial derivatives of a function. It helps us determine the curvature of the surface at the critical points. For functions of two variables like ours, the Hessian matrix is a 2x2 matrix.

Specifically, the Hessian matrix for our function is composed of the second partial derivatives:
  • \( \frac{\partial^2 f}{\partial x^2} = -10.4420 \)
  • \( \frac{\partial^2 f}{\partial y^2} = -0.1874 \)
  • \( \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} = -0.4023 \)

We form the Hessian matrix as:
\ \[ \begin{bmatrix} -10.4420 & -0.4023 \ -0.4023 & -0.1874 \end{bmatrix} \].

Analyzing this matrix helps us use the second derivative test to characterize the nature of the critical point.
Second Derivative Test
The second derivative test is a method for determining whether a critical point is a local maximum, minimum, or a saddle point. It uses the determinant of the Hessian matrix. For a critical point in a two-variable function \( f(x, y) \), the test involves calculating the determinant \( D \) of the Hessian matrix:

\ \[ D = \left| \begin{array}{cc} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \ \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2} \end{array} \right| \].

In our case, \ \[ D = (-10.4420)(-0.1874) - (-0.4023)^2 \].

The value of \( D \) indicates the nature of the critical point:
  • If \( D > 0 \) and \( \frac{\partial^2 f}{\partial x^2} < 0 \), it's a local maximum.
  • If \( D > 0 \) and \( \frac{\partial^2 f}{\partial x^2} > 0 \), it's a local minimum.
  • If \( D < 0 \), the point is a saddle point.
Given that \( D > 0 \) and \( \frac{\partial^2 f}{\partial x^2} < 0 \), the critical point \((x, y) = (28.74, 61.76)\) is a local maximum, indicating the most favorable conditions for egg production.
Catolaccus grandis
Catolaccus grandis is an important biological control agent used to manage boll weevil populations, which are harmful to cotton crops. This small parasitic wasp lays its eggs on the larvae of boll weevils, and these larvae eventually feed on and kill their hosts.

Understanding the optimal conditions for the reproduction of Catolaccus grandis is crucial for maximizing its effectiveness in pest management. The function provided in the exercise models the number of eggs produced by \( C. grandis \) under certain environmental conditions. By using calculus techniques like finding critical points and conducting a second derivative test, we can determine the best conditions to increase egg production.

This process not only improves the efficiency of \( C. grandis \) as a biological control agent but also contributes to sustainable agricultural practices by reducing the reliance on chemical pesticides. With such data-driven insights, farmers can enhance the resilience of crops against pests in an environmentally friendly way.

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