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Magazine Chapter 13: Problem 8
Find all stationary points \(\left(f_{x}=f_{y}=0\right)\) in \(1-16 .\) Separate minimum from maximum from saddle point. Test \(13 \mathrm{~K}\) applies to \(a=f_{x x}, b=f_{x y}, c=f_{y y}\) \((x+y)^{2}+(x+2 y-6)^{2}\)
Short Answer
Expert verified
The stationary point is \((-1.5, 4.5)\) and it is a local minimum.
Step by step solution
01
Identify the function given
The function given in the problem is \[f(x, y) = (x+y)^2 + (x+2y-6)^2.\]
02
Find first partial derivatives
The first partial derivatives of the function are needed to find the stationary points:- The partial derivative with respect to \(x\) is: \[f_x = \frac{\partial}{\partial x}[(x+y)^2 + (x+2y-6)^2] = 2(x+y) + 2(x+2y-6).\] Simplifying, we have \[f_x = 4x + 4y - 12.\]- The partial derivative with respect to \(y\) is: \[f_y = \frac{\partial}{\partial y}[(x+y)^2 + (x+2y-6)^2] = 2(x+y) + 4(x+2y-6).\] Simplifying, we have \[f_y = 6y + 2x - 24.\]
03
Set partial derivatives equal to zero (Stationary points condition)
Set the partial derivatives equal to zero to find the critical points:1) \(4x + 4y - 12 = 0\)2) \(6y + 2x - 24 = 0\).Solve this system of equations simultaneously.
04
Solve the system of equations
Solve the system of equations:1) \(4x + 4y = 12\) => \(x + y = 3\) (divide the entire equation by 4)2) \(2x + 6y = 24\) => \(x + 3y = 12\) (divide the entire equation by 2)Subtract equation (1) from equation (2): - \(x + 3y = 12\)- \(x + y = 3\)Gives \(2y = 9\) so \(y = 4.5\).Substitute \(y = 4.5\) into \(x + y = 3\), we have:- \(x + 4.5 = 3\)Solving for \(x\), we get \(x = -1.5\).Thus, the critical point is \((-1.5, 4.5)\).
05
Find second partial derivatives and calculate determinant for test
Find the second partial derivatives to use the second derivative test:\[\begin{align*} f_{xx} &= \frac{\partial^2 f}{\partial x^2} = 4, \ f_{yy} &= \frac{\partial^2 f}{\partial y^2} = 6, \ f_{xy} &= \frac{\partial^2 f}{\partial x \partial y} = 4.\end{align*}\]Now, calculate the determinant \(D\) at the stationary point \[D = f_{xx}f_{yy} - (f_{xy})^2 = 4 \cdot 6 - 4^2 = 24 - 16 = 8.\]
06
Classify the stationary point
Since \(D > 0\) and \(f_{xx} > 0\) at \((-1.5, 4.5)\), this stationary point is a local minimum.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivative
Partial derivatives help us examine how a function changes as we vary one variable while keeping others constant. In our exercise, we take partial derivatives with respect to both \(x\) and \(y\) for the function \(f(x, y) = (x+y)^2 + (x+2y-6)^2\). We need these derivatives to locate where the slopes in both directions are zero, indicating a stationary point.
- For the partial derivative with respect to \(x\), denoted as \(f_x\), the computation involves treating \(y\) as constant and differentiating with respect to \(x\). We get \(f_x = 4x + 4y - 12\).
- Similarly, for the partial derivative with respect to \(y\), represented as \(f_y\), \(x\) is treated as constant while differentiating with respect to \(y\). We find \(f_y = 6y + 2x - 24\).
Second Derivative Test
Once we find a stationary point, we use the second derivative test to classify its nature鈥攚hether it's a local minimum, maximum, or a saddle point. The test involves computing second partial derivatives and forming a determinant \(D\).For our function, we calculate:
- \(f_{xx} = 4\)
- \(f_{yy} = 6\)
- \(f_{xy} = f_{yx} = 4\)
- If \(D > 0\) and \(f_{xx} > 0\), the stationary point is a local minimum.
- If \(D > 0\) and \(f_{xx} < 0\), it is a local maximum.
- If \(D < 0\), it's a saddle point.
Critical Points
Critical points of a function occur where its partial derivatives are equal to zero. These are potential spots for local minima, maxima, or saddle points. For our function, we solved a system of equations derived from the conditions \(f_x = 0\) and \(f_y = 0\):
- \(4x + 4y = 12\), simplifying to \(x + y = 3\)
- \(2x + 6y = 24\), simplifying to \(x + 3y = 12\)
Local Minimum
A local minimum is a point where the function value is lower than all nearby points. To confirm a local minimum, the second derivative test provides a clear criteria: both \(D > 0\) and \(f_{xx} > 0\).In our scenario, after extensive calculations, we reached \(D = 8\) and \(f_{xx} = 4\). Since both conditions are met, the critical point \((-1.5, 4.5)\) is indeed a local minimum. This means in the vicinity of \((-1.5, 4.5)\), the value of our function \((x+y)^2 + (x+2y-6)^2\) will not be higher than any other points close by.Understanding where a local minimum occurs is crucial, especially in optimization problems, as it indicates a potential point of lowest cost or minimal value.
