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

Short Answer

Expert verified
The critical points are \((0, 0)\) and \((18.503, 4.7226)\), both are saddle points.

Step by step solution

01

Calculate the Partial Derivatives

To find the critical points of the function \( h(w, z) = 0.6w^{2} + 1.3z^{3} - 4.7wz \), calculate the partial derivatives with respect to \( w \) and \( z \). First, the partial derivative with respect to \( w \):\[ h_w = \frac{\partial}{\partial w}(0.6w^{2} + 1.3z^{3} - 4.7wz) = 1.2w - 4.7z. \]Next, the partial derivative with respect to \( z \):\[ h_z = \frac{\partial}{\partial z}(0.6w^{2} + 1.3z^{3} - 4.7wz) = 3.9z^{2} - 4.7w. \]
02

Set the Partial Derivatives Equal to Zero

To find the critical points, set the partial derivatives \( h_w \) and \( h_z \) equal to zero and solve for \( w \) and \( z \):1. Solve \( 1.2w - 4.7z = 0 \).2. Solve \( 3.9z^{2} - 4.7w = 0 \).
03

Solve the System of Equations

From the equations:1. \( 1.2w = 4.7z \) implies \( w = \frac{4.7}{1.2}z \).2. Substitute \( w = \frac{4.7}{1.2}z \) into \( 3.9z^{2} = 4.7w \):\[ 3.9z^{2} = 4.7\left(\frac{4.7}{1.2}z\right), \]\[ 3.9z^{2} = \frac{22.09}{1.2}z, \]\[ 3.9z^{2} = 18.4083z. \]3. Factor the equation: \( 3.9z(z - 4.7226) = 0 \).4. Solutions: \( z = 0 \) or \( z \approx 4.7226 \).5. For \( z = 0 \): \( w = \frac{4.7}{1.2} \times 0 = 0 \), thus the critical point is \((0, 0)\).6. For \( z \approx 4.7226 \): \( w = \frac{4.7}{1.2} \times 4.7226 \approx 18.503 \), thus the point is \((18.503, 4.7226)\).
04

Classify the Critical Points Using the Second Derivative Test

Calculate the second partial derivatives:\[ h_{ww} = \frac{\partial^2 h}{\partial w^2} = 1.2, \]\[ h_{zz} = \frac{\partial^2 h}{\partial z^2} = 7.8z, \]\[ h_{wz} = \frac{\partial^2 h}{\partial w \partial z} = -4.7. \]For the critical point \((0, 0)\), the Hessian determinant \( D = h_{ww}h_{zz} - (h_{wz})^2 \) is:\[ D = 1.2 \cdot 7.8 \cdot 0 - (-4.7)^2 = -22.09, \]which is negative, indicating a saddle point.For \((18.503, 4.7226)\), \( h_{zz} \approx 36.8617 \):\[ D = 1.2 \cdot 36.8617 - (-4.7)^2 = -22.09, \]which is negative, so it is also a saddle point.

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Key Concepts

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

Partial Derivatives
When dealing with multivariable functions, understanding how variables change independently is crucial. Partial derivatives allow us to explore this by examining the rate of change of a function concerning one variable while keeping other variables constant.
For example, given a function \( h(w, z) = 0.6w^2 + 1.3z^3 - 4.7wz \), you can find the partial derivative with respect to \( w \), noted as \( h_w \), by treating \( z \) as a constant. The same principle applies when finding the partial derivative with respect to \( z \), known as \( h_z \).
Calculating partial derivatives helps identify critical points, where the function's slope is temporarily zero. These points occur where \( h_w = 0 \) and \( h_z = 0 \), offering vital information about the function's behavior, like whether it reaches a peak, valley, or saddle point.
Hessian Determinant
The Hessian matrix plays a fundamental role in determining the nature of critical points. It is a square matrix of second-order partial derivatives, structured to provide deeper insight into the behavior around critical points.
For our function, the Hessian matrix \( H \) is constructed from the second-order partial derivatives: \( h_{ww} \), \( h_{zz} \), and \( h_{wz} \). Specifically:
  • \( h_{ww} = \frac{\partial^2 h}{\partial w^2} = 1.2 \)
  • \( h_{zz} = \frac{\partial^2 h}{\partial z^2} = 7.8z \)
  • \( h_{wz} = \frac{\partial^2 h}{\partial w \partial z} = -4.7 \)
The Hessian determinant \( D \) is calculated as \( h_{ww}h_{zz} - (h_{wz})^2 \). This determinant helps us classify the nature of critical points:
  • If \( D > 0 \) and all second partial derivatives are positive, the point is a local minimum.
  • If \( D > 0 \) and at least one second partial derivative is negative, the point is a local maximum.
  • If \( D < 0 \), the point is a saddle point, indicating neither a max nor a min.
Second Derivative Test
To classify the critical points identified using partial derivatives, we employ the Second Derivative Test. This test uses the Hessian matrix to evaluate the nature of these points by examining the concavity of the function in multiple dimensions.
The outcome of this test is reliant on the Hessian determinant \( D \):
  • If \( D > 0 \), the point may be an extremum (either maximum or minimum depending on whether \( h_{ww} \) is positive or negative).
  • If \( D < 0 \), it indicates a saddle point, a point where the function has traits of both increasing and decreasing tendencies.
  • If \( D = 0 \), the test is inconclusive, and further analysis is needed to determine the point's nature.

In our exercise, both critical points \((0,0)\) and \((18.503, 4.7226)\) resulted in a negative Hessian determinant, classifying them as saddle points. This suggests these points possess characteristics of neither absolute highs nor lows but rather mixed behaviors in different directions.

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Most popular questions from this chapter

Locate and classify any critical points. $$ f(x, y)=3 x^{2}-x^{3}+12 y^{2}-8 y^{3}+60 $$

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