Problem 3
Question
Suppose \(f(x, y)=x y-a x-b y\). (A) How many local minimum points does \(f\) have in \(\mathbf{R}^{2} ?\) (The answer is an integer). (B) How many local maximum points does \(f\) have in \(\mathbf{R}^{2} ?\) (C) How many saddle points does \(f\) have in \(\mathbf{R}^{2} ?\)
Step-by-Step Solution
Verified Answer
The function \(f(x, y) = xy - ax - by\) has:
(A) 0 local minimum points.
(B) 0 local maximum points.
(C) 1 saddle point.
1Step 1: Calculate first partial derivatives
First, we should calculate the first partial derivatives of the function f(x, y) with respect to x and y:
\(\frac{\partial f}{\partial x} = y- a\)
\(\frac{\partial f}{\partial y} = x - b\)
2Step 2: Set partial derivatives equal to 0 to find critical points
Next, we set the partial derivatives equal to 0 and solve for x and y:
\(y - a = 0 \Rightarrow y = a\)
\(x - b = 0 \Rightarrow x = b\)
Therefore, the only critical point is (b, a).
3Step 3: Calculate second partial derivatives
Now, let's calculate the second partial derivatives:
\(\frac{\partial^2 f}{\partial x^2} = 0\)
\(\frac{\partial^2 f}{\partial y^2} = 0\)
\(\frac{\partial^2 f}{\partial x\partial y} = \frac{\partial^2 f}{\partial y\partial x} = 1\)
4Step 4: Use the second derivative test
To determine the classification of the critical point, we use the second derivative test:
Let \(D(x, y) = \frac{\partial^2 f}{\partial x^2} \frac{\partial^2 f}{\partial y^2} - (\frac{\partial^2 f}{\partial x \partial y})^2\)
\(D(b, a) = (0)(0) - (1)^2=-1\)
Since \(D(b, a) < 0\), the critical point (b, a) is a saddle point.
5Step 5: Answer
(A) There are 0 local minimum points.
(B) There are 0 local maximum points.
(C) There is 1 saddle point.
Key Concepts
Critical PointsPartial DerivativesSecond Derivative TestSaddle Point
Critical Points
Critical points are special locations on the graph of a function where the gradient is zero or undefined. For a function of two variables like \(f(x, y)\), these points occur where both of the first partial derivatives equal zero.
To find these points, you calculate the partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\). Once you've found those, set them to zero and solve the system of equations that results.
In this exercise, the critical point is found at \((b, a)\) since both \(\frac{\partial f}{\partial x} = y - a\) and \(\frac{\partial f}{\partial y} = x - b\) equals zero at this pair of \(x\) and \(y\) values.
To find these points, you calculate the partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\). Once you've found those, set them to zero and solve the system of equations that results.
In this exercise, the critical point is found at \((b, a)\) since both \(\frac{\partial f}{\partial x} = y - a\) and \(\frac{\partial f}{\partial y} = x - b\) equals zero at this pair of \(x\) and \(y\) values.
Partial Derivatives
Partial derivatives are a fundamental tool in multivariable calculus. They measure the rate of change of a function with respect to one variable while keeping others constant. For the function \(f(x, y) = x y - a x - b y\), the first partial derivatives with respect to \(x\) and \(y\) are crucial for identifying critical points.
Here, the partial derivative with respect to \(x\) is \(\frac{\partial f}{\partial x} = y - a\). This indicates how \(f\) changes as \(x\) changes, holding \(y\) constant. Similarly, the derivative \(\frac{\partial f}{\partial y} = x - b\) represents the change in \(f\) with variations in \(y\).
These derivatives help locate the critical points by setting them to zero and solving for \(x\) and \(y\).
Here, the partial derivative with respect to \(x\) is \(\frac{\partial f}{\partial x} = y - a\). This indicates how \(f\) changes as \(x\) changes, holding \(y\) constant. Similarly, the derivative \(\frac{\partial f}{\partial y} = x - b\) represents the change in \(f\) with variations in \(y\).
These derivatives help locate the critical points by setting them to zero and solving for \(x\) and \(y\).
Second Derivative Test
The second derivative test helps classify critical points as local minima, local maxima, or saddle points. After finding the critical points, the next step is to examine the second partial derivatives.
For the function \(f(x, y)\), second partial derivatives are given by \(\frac{\partial^2 f}{\partial x^2}\), \(\frac{\partial^2 f}{\partial y^2}\), and the mixed partial derivative \(\frac{\partial^2 f}{\partial x\partial y}\). These are used to compute \(D(x, y)\), where:
If \(D < 0\), the critical point is a saddle point. If \(D > 0\) and both second partial derivatives are positive, it is a local minimum, and if negative, a local maximum. In this function's case, \(D(b, a) = -1 < 0\), indicating a saddle point.
For the function \(f(x, y)\), second partial derivatives are given by \(\frac{\partial^2 f}{\partial x^2}\), \(\frac{\partial^2 f}{\partial y^2}\), and the mixed partial derivative \(\frac{\partial^2 f}{\partial x\partial y}\). These are used to compute \(D(x, y)\), where:
- \(D(x, y) = \frac{\partial^2 f}{\partial x^2} \frac{\partial^2 f}{\partial y^2} - (\frac{\partial^2 f}{\partial x \partial y})^2\).
If \(D < 0\), the critical point is a saddle point. If \(D > 0\) and both second partial derivatives are positive, it is a local minimum, and if negative, a local maximum. In this function's case, \(D(b, a) = -1 < 0\), indicating a saddle point.
Saddle Point
A saddle point is a type of critical point that is neither a local maximum nor a local minimum. Instead, it is a location on the graph where the surface has a saddle-like shape—concave in one direction and convex in the perpendicular direction.
Using the second derivative test, a saddle point is confirmed if \(D < 0\), as shown in the example problem for the function \(f(x, y)\). At the critical point \((b, a)\), the calculation of \(D(b, a) = -1\) indicates this is indeed a saddle point.
This classification means that near the point \((b, a)\), the function \(f(x, y)\) behaves differently along different paths. Some paths will show an increase while others will show a decrease, reflecting the "saddle" structure.
Using the second derivative test, a saddle point is confirmed if \(D < 0\), as shown in the example problem for the function \(f(x, y)\). At the critical point \((b, a)\), the calculation of \(D(b, a) = -1\) indicates this is indeed a saddle point.
This classification means that near the point \((b, a)\), the function \(f(x, y)\) behaves differently along different paths. Some paths will show an increase while others will show a decrease, reflecting the "saddle" structure.
Other exercises in this chapter
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