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Magazine Chapter 14: Problem 5
Find all the local maxima, local minima, and saddle points of the functions in Exercises \(1-30 .\) $$ f(x, y)=2 x y-x^{2}-2 y^{2}+3 x+4 $$
Short Answer
Expert verified
The local maximum is at \((3, \frac{3}{2})\).
Step by step solution
01
Find the First Partial Derivatives
To find the critical points, we begin by finding the first partial derivatives of the function. The function given is \[ f(x, y) = 2xy - x^2 - 2y^2 + 3x + 4. \]Compute \(f_x\):\[ f_x = \frac{\partial}{\partial x}(2xy - x^2 - 2y^2 + 3x + 4) = 2y - 2x + 3. \]Then compute \(f_y\):\[ f_y = \frac{\partial}{\partial y}(2xy - x^2 - 2y^2 + 3x + 4) = 2x - 4y. \]
02
Set the Partial Derivatives to Zero
Next, we set the first partial derivatives equal to zero to find the critical points:1. \(2y - 2x + 3 = 0\)2. \(2x - 4y = 0\).
03
Solve the System of Equations
We will solve the system of equations from step 2:1. From the equation \(2x - 4y = 0\), we can simplify it to \(x = 2y\).2. Substitute \(x = 2y\) into the equation \(2y - 2x + 3 = 0\): \(2y - 2(2y) + 3 = 0\) \(-2y + 3 = 0\) \(y = \frac{3}{2}\).3. Substitute \(y = \frac{3}{2}\) back into \(x = 2y\): \(x = 2 \times \frac{3}{2} = 3\).4. Thus, the critical point is \((x, y) = (3, \frac{3}{2})\).
04
Find the Second Partial Derivatives
To classify the critical point, we need the second partial derivatives:- \(f_{xx} = \frac{\partial}{\partial x}(2y - 2x + 3) = -2\)- \(f_{yy} = \frac{\partial}{\partial y}(2x - 4y) = -4\)- \(f_{xy} = \frac{\partial}{\partial y}(2y - 2x + 3) = 2\)
05
Use the Second Derivative Test
We apply the second derivative test using:\[ D = f_{xx}f_{yy} - (f_{xy})^2. \]Substituting gives:\[ D = (-2)(-4) - (2)^2 = 8 - 4 = 4. \]Since \(D > 0\) and \(f_{xx} < 0\), \((3, \frac{3}{2})\) is a local maximum.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Critical Points
Critical points in multivariable calculus are points where the gradient of a function is zero, meaning the function has stationary behavior there鈥攅ither a minimum, maximum, or a saddle point. For a function of two variables, such as \(f(x, y)=2xy-x^2-2y^2+3x+4\), these points are found by setting the first partial derivatives to zero and solving for \(x\) and \(y\).
- If both partial derivatives, \(f_x\) and \(f_y\), are zero at a point, then that point is a critical point.
- The function behaves like a turning point, but further tests are needed to know its exact nature.
Partial Derivatives
Partial derivatives are used to find the rate of change of a function in one specific direction while keeping the other variables constant. They are the building blocks for finding critical points in a multivariable function. For the function \(f(x, y) = 2xy - x^2 - 2y^2 + 3x + 4\), we find the partial derivatives with respect to \(x\) and \(y\).
- \(f_x = 2y - 2x + 3\) represents change moving along the x-direction.
- \(f_y = 2x - 4y\) represents change along the y-direction.
Second Derivative Test
After finding critical points, the second derivative test helps determine their nature. This test evaluates whether a critical point is a local minimum, maximum, or a saddle point. For a two-variable function, second partial derivatives and the discriminant \(D\) play crucial roles.
- Compute \(f_{xx}, f_{yy},\) and \(f_{xy}\) to set up the test.
- The discriminant \(D\ = f_{xx}f_{yy} - (f_{xy})^2\).
Local Maxima and Minima
Local maxima and minima are types of critical points where the function achieves a highest or lowest value in a small surrounding region, respectively.
- A point is a local maximum if, near that point, the function values are lower than at that point.
- A point is a local minimum if the surrounding function values are higher.
- If \(D > 0\) and \(f_{xx} > 0\), the point is a local minimum.
- If \(D > 0\) and \(f_{xx} < 0\), the point is a local maximum, like in our solved example.
Saddle Points
Saddle points are unique critical points where a function neither reaches a local maximum nor minimum鈥攊t "saddles" between increasing and decreasing directions.
- At such a point, the function curves upward in one direction and downward in another.
- The discriminant \(D\) for saddle points is \(D < 0\).
