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Chapter 8: Problem 1

Locate and classify any critical points. $$ R(k, m)=3 k^{2}-2 k m-20 k+3 m^{2}-4 m+60 $$

Short Answer

Expert verified
The critical point is \((k, m) = (4, 2)\), which is a local minimum.

Step by step solution

01

Understand the Problem

We need to locate and classify any critical points for the function \( R(k, m) = 3k^2 - 2km - 20k + 3m^2 - 4m + 60 \). Critical points occur where the first partial derivatives are equal to zero.
02

Find Partial Derivatives

Compute the partial derivative of \( R \) with respect to \( k \):\[ R_k = \frac{\partial R}{\partial k} = 6k - 2m - 20 \]Compute the partial derivative of \( R \) with respect to \( m \):\[ R_m = \frac{\partial R}{\partial m} = -2k + 6m - 4 \].
03

Set Partial Derivatives to Zero

Set the partial derivatives equal to zero to find critical points:\[ 6k - 2m - 20 = 0 \]\[ -2k + 6m - 4 = 0 \].
04

Solve the System of Equations

Solve the system of equations:1. \( 6k - 2m = 20 \)2. \( -2k + 6m = 4 \)From equation (1): \( 6k = 2m + 20 \), which simplifies to \( k = \frac{m + 10}{3} \).Substitute into equation (2):\[ -2\left(\frac{m + 10}{3}\right) + 6m = 4 \]Multiply through by 3 to eliminate the fraction:\[ -2(m + 10) + 18m = 12 \]\[ -2m - 20 + 18m = 12 \]Combine like terms:\[ 16m - 20 = 12 \]\[ 16m = 32 \]\[ m = 2 \].Substitute \( m = 2 \) back into \( k = \frac{m + 10}{3} \):\[ k = \frac{2 + 10}{3} = 4 \].So, the critical point is \( (k, m) = (4, 2) \).
05

Classify the Critical Point Using the Second Derivative Test

Compute the second partial derivatives:\[ R_{kk} = \frac{\partial^2 R}{\partial k^2} = 6, \quad R_{mm} = \frac{\partial^2 R}{\partial m^2} = 6, \quad R_{km} = \frac{\partial^2 R}{\partial k \partial m} = -2 \].Find the determinant of the Hessian matrix:\[ H = \begin{vmatrix} R_{kk} & R_{km} \ R_{km} & R_{mm} \end{vmatrix} = \begin{vmatrix} 6 & -2 \ -2 & 6 \end{vmatrix} = (6)(6) - (-2)(-2) = 36 - 4 = 32 \].Since \( H > 0 \) and \( R_{kk} > 0 \), the critical point \( (4, 2) \) is a local minimum.

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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 special points on a function where the slope is either zero or undefined. These points are important because they can indicate where a function reaches a maximum, minimum, or a saddle point. To find critical points for a function of two variables, like in the exercise, you need to find where the first partial derivatives of the function are both zero.
  • Critical points provide insights into the behavior of the function.
  • They are potential locations for local maxima, minima, or saddle points.
Once you spot a critical point, further tests are needed to classify it, which is where other multivariable calculus tools come into play.
Partial Derivatives
Partial derivatives are derivatives of functions with several variables, where you differentiate with respect to one variable at a time, treating the other variables as constants. In the context of the exercise, we differentiate the function \( R(k, m) = 3k^2 - 2km - 20k + 3m^2 - 4m + 60 \) with respect to each variable:
  • \( R_k = \frac{\partial R}{\partial k} = 6k - 2m - 20 \)
  • \( R_m = \frac{\partial R}{\partial m} = -2k + 6m - 4 \)
These derivatives help us find critical points by setting them equal to zero. They essentially tell us how the function changes as each variable is varied, leading us to potential locations of interest.
Second Derivative Test
The second derivative test is a method used to determine the nature of critical points in multivariable functions. After finding critical points, this test helps classify them as local minima, maxima, or saddle points. Once you calculate the second derivatives and find the Hessian Matrix, you determine:
  • If the determinant of the Hessian matrix (\( H \)) > 0 and \( R_{kk} > 0 \), the critical point is a local minimum.
  • If \( H > 0 \) and \( R_{kk} < 0 \), it's a local maximum.
  • If \( H < 0 \), the point is a saddle point.
  • If \( H = 0 \), the test is inconclusive.
This test is essential for understanding the shape and behavior of surfaces described by multivariable functions.
Hessian Matrix
The Hessian Matrix is a square matrix of second-order partial derivatives of a scalar-valued function. It plays a crucial role in the second derivative test. The Hessian for our function is given as: \[ H = \begin{vmatrix} R_{kk} & R_{km} \R_{mk} & R_{mm} \end{vmatrix} \] Where for our specific function \( R(k, m) \), you had:
  • \( R_{kk} = 6 \)
  • \( R_{mm} = 6 \)
  • \( R_{km} = -2 \)
Calculating the determinant provides insight into whether you've got a saddle point, minimum or maximum at the critical point. In this example, the determinant was \( 32 \), indicating a local minimum. Understanding the Hessian Matrix helps in visually and mathematically comprehending the behavior of multivariable functions at their critical points.

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Most popular questions from this chapter

a. Write the Lagrange system of partial derivative equations. b. Locate the optimal point of the constrained system. c. Identify the optimal point as either a maximum point or a minimum point. $$ \left\\{\begin{array}{l} \text { optimize } f(a, b)=a b \\ \text { subject to } g(a, b)=a^{2}+b^{2}=90 \end{array}\right. $$

For each of the tables a. Write \(S S E\) as a multivariable function \(f\) of \(a\) and \(b\) for the best- fitting line \(y=a x+b\). b. Write expressions for \(\frac{\partial f}{\partial a}, \frac{\partial f}{\partial b},\) and the determinant of the second partials matrix. c. Locate the minimum point of \(f(a, b)\). d. Use the results of part \(c\) to write the linear function that best fits the data. $$ \begin{array}{|c|c|c|c|} \hline x & 1 & 6 & 12 \\ \hline y & 7 & 11 & 19 \\ \hline \end{array} $$

Locate and classify any critical points. $$ h(w, z)=0.6 w^{2}+1.3 z^{3}-4.7 w z $$

Mulch Profit A nursery sells mulch by the truckload. Bark mulch sells for \(\$ b\) per load, and pine straw sells for \(\$ p\) per load. The nursery's average weekly profit from the sale of these two types of mulch can be modeled as \(P(p, b)=144 p-3 p^{2}-p b-2 b^{2}+120 b+35\) dollars a. How much should the nursery charge for each type of mulch to maximize the weekly profit from the sale of mulch? b. Explain how the Determinant Test verifies that the result in part \(a\) gives the maximum profit. c. What is the maximum weekly profit from the sales of these two types of mulch?

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