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Magazine Chapter 13: Problem 1
Find the critical points, if any, of \(F\). a. \(\quad F(x, y)=2 x+5 y+7\) b. \(\quad F(x, y)=x^{2}+4 x y+3 y^{2}\) c. \(\quad F(x, y)=x^{3}(1-x)+y\) d. \(\quad F(x, y)=x y(1-x y)\) e. \(\quad F(x, y)=\left(x-x^{2}\right)\left(y-y^{2}\right) \quad\) f. \(\quad F(x, y)=\frac{x}{y}\) g. \(\quad F(x, y)=e^{x+y}\) h. \(\quad F(x, y)=\sin (x+y)\) \(\begin{array}{lll}\text { i. } & F(x, y)=\frac{x^{2}}{1+y^{2}} & \text { j. } \quad F(x, y)\end{array}=\cos x \sin y\)
Short Answer
Expert verified
Critical points: a) None, b) (0,0), c) x=0 or x=3/4, d) (0,0) & 2xy=1 curve, e) (0,0), (0,1), (1,0), (1,1), (1/2, 1/2), f) None, g) None, h) x+y=蟺/2+n蟺, i) (0,y), j) x=n蟺, y=蟺/2+l蟺.
Step by step solution
01
Understand Critical Points
Critical points of a multivariable function occur where the gradient (partial derivatives set equal to zero) is zero or undefined. To find these, we calculate the partial derivatives of the function with respect to each variable and set them to zero.
02
Solve Part a
The function is given by \( F(x, y) = 2x + 5y + 7 \). Calculate the partial derivatives: \( F_x = 2 \) and \( F_y = 5 \). Both are constants, so there are no critical points because neither partial derivative can be zero unless the constant value was zero.
03
Solve Part b
The function is \( F(x, y) = x^2 + 4xy + 3y^2 \). Calculate the partial derivatives: \( F_x = 2x + 4y \) and \( F_y = 4x + 6y \). Set them to zero, \( 2x + 4y = 0 \) and \( 4x + 6y = 0 \). Solving these equations, we find \( x = 0 \) and \( y = 0 \). Thus, there is one critical point at (0, 0).
04
Solve Part c
The function is \( F(x, y) = x^3(1-x) + y \). Calculate the partial derivatives: \( F_x = 3x^2(1-x) - x^3 \) and \( F_y = 1 \). The partial derivative \( F_y = 1 \) means there are no extrema in the y-direction. Setting \( 3x^2(1-x) - x^3 = x^2(3 - 4x) = 0 \) yields \( x = 0 \) or \( x = \frac{3}{4} \). Hence, critical points occur where \( x = 0 \) or \( x = \frac{3}{4} \).
05
Solve Part d
The function is \( F(x, y) = xy(1 - xy) \). Calculate the partial derivatives: \( F_x = y(1 - 2xy) \) and \( F_y = x(1 - 2xy) \). Setting these to zero, \( y(1 - 2xy) = 0 \) implies \( y = 0 \) or \( 2xy = 1 \). Similarly, \( x(1 - 2xy) = 0 \) implies \( x = 0 \) or \( 2xy = 1 \). Solving these, we find critical points at (0, 0) and points on the curve given by \( 2xy = 1 \), such as \( (1, \frac{1}{2}) \) and \( (\frac{1}{2}, 1) \).
06
Solve Part e
The function is \( F(x, y) = (x-x^2)(y-y^2) \). Calculate the partial derivatives: \( F_x = (1-2x)(y-y^2) \) and \( F_y = (x-x^2)(1-2y) \). Setting these to zero, \((1-2x)(y-y^2) = 0\) implies \( x = \frac{1}{2} \) or \( y = 0 \), \( y = 1 \). Similarly, \((x-x^2)(1-2y) = 0\) implies \( y = \frac{1}{2} \) or \( x = 0 \), \( x = 1 \). This calculation gives potential critical points at (0, 0), (0, 1), (1, 0), (1, 1), (\frac{1}{2}, \frac{1}{2}).
07
Solve Part f
The function is \( F(x, y) = \frac{x}{y} \). Calculate the partial derivatives: \( F_x = \frac{1}{y} \) and \( F_y = -\frac{x}{y^2} \). There are no critical points because \( F_x \) can never be zero unless \( y = \infty \), and \( F_y \) goes to infinity as \( y \to 0 \).
08
Solve Part g
The function is \( F(x, y) = e^{x+y} \). The partial derivatives are \( F_x = e^{x+y} \) and \( F_y = e^{x+y} \). No partial derivative can be zero because the exponential function cannot be zero. Therefore, there are no critical points.
09
Solve Part h
The function is \( F(x, y) = \sin(x+y) \). The partial derivatives are \( F_x = \cos(x+y) \) and \( F_y = \cos(x+y) \). Set \( \cos(x+y) = 0 \), leading to \( x+y = \frac{\pi}{2} + n\pi \), where \( n \) is an integer. Thus, critical points occur on lines given by \( x+y = \frac{\pi}{2} + n\pi \).
10
Solve Part i
The function is \( F(x, y) = \frac{x^2}{1+y^2} \). Calculate the partial derivatives: \( F_x = \frac{2x}{1+y^2} \) \( F_y = -\frac{x^2(2y)}{(1+y^2)^2} \). Setting \( F_x = 0 \) gives \( x = 0 \), and \( F_y = 0 \) unless \( x^2 = 0 \). Hence, the only critical point is \( (0, y) \) for any \( y \).
11
Solve Part j
The function is \( F(x, y) = \cos x \sin y \). Partial derivatives are \( F_x = -\sin x \sin y \) and \( F_y = \cos x \cos y \). Both must be zero: \(-\sin x \sin y = 0\) implies \( x = n\pi \) or \( y = m\pi \). For \( \cos x \cos y = 0 \), solve \( x = \frac{\pi}{2} + k\pi \) or \( y = \frac{\pi}{2} + l\pi \). Combining, the critical points occur at intersections of these values.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gradient
In multivariable calculus, the gradient is an essential concept that represents the direction and rate of fastest increase of a function. For a function with several variables, such as two variables, the gradient is a vector composed of partial derivatives with respect to each variable.
- If we have a function \( F(x, y) \), the gradient \( abla F \) is given by the vector \( (F_x, F_y) \), where \( F_x \) is the partial derivative of \( F \) with respect to \( x \), and \( F_y \) is the partial derivative with respect to \( y \).
- The gradient helps in finding critical points by indicating locations where these derivatives become zero or are undefined.
- In a two-dimensional space, the gradient points in the direction of the steepest ascent.
Partial Derivatives
Partial derivatives are the building blocks of multivariable calculus, allowing you to analyze the behavior of functions with two or more variables with respect to one variable at a time. When working with a function \( F(x, y) \), you may find the following:
- \( F_x \), the partial derivative with respect to \( x \), treats \( y \) as a constant and determines how \( F \) changes as \( x \) changes.
- \( F_y \), the partial derivative with respect to \( y \), similarly treats \( x \) as a constant.
- They are used to calculate the gradient, which is crucial for finding critical points of multivariable functions.
- They provide insights into the variation of the function along different axes.
Extrema in Functions
In calculus, extrema refer to various types of important points where a function reaches its maximum or minimum values. When dealing with functions of multiple variables, identifying these points can be more complex than in single-variable functions.
- Critical Points: These are points where the gradient is zero or undefined, potentially indicating local maxima, minima, or saddle points.
- Local vs Global Extrema: Local extrema are where a function has a peak or valley relative to nearby points. Global extrema refer to the absolute highest or lowest points across the entire domain.
Solving Equations for Critical Points
Finding critical points in multivariable calculus involves setting partial derivatives equal to zero and solving the resulting system of equations. This process will illuminate the behavior of the function at specific points.
- Start by computing the partial derivatives \( F_x \) and \( F_y \) of the function \( F(x, y) \).
- Set each partial derivative equal to zero, forming a system of equations that represents potential critical points.
- Use algebraic methods to solve this system, which might involve solving for one variable at a time or employing substitution or elimination techniques.
- Once solutions are found, these candidate points need further analysis to determine their nature (e.g., maximum, minimum, or saddle point).
