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Chapter 14: Problem 31

In Exercises \(31-38\) , find the absolute maxima and minima of the functions on the given domains. \(f(x, y)=2 x^{2}-4 x+y^{2}-4 y+1\) on the closed triangular plate bounded by the lines \(x=0, y=2, y=2 x\) in the first quadrant

Short Answer

Expert verified
The absolute maximum is 1 at (0,0) and the minimum is -3 at (0,2) and (1,2).

Step by step solution

01

Determine the Critical Points

To find the absolute maxima and minima, we first need to identify the critical points of the function. Compute the partial derivatives.The partial derivative with respect to x is: \[ f_x = \frac{\partial}{\partial x}(2x^2 - 4x + y^2 - 4y + 1) = 4x - 4 \]Set this equal to zero: \[ 4x - 4 = 0 \quad \Rightarrow \quad x = 1 \]The partial derivative with respect to y is: \[ f_y = \frac{\partial}{\partial y}(2x^2 - 4x + y^2 - 4y + 1) = 2y - 4 \]Set this equal to zero: \[ 2y - 4 = 0 \quad \Rightarrow \quad y = 2 \]Thus, the critical point is \((1, 2)\). Check if this point falls within the triangular region.
02

Evaluate on Boundary Lines

The boundary lines of the triangle are \(x = 0\), \(y = 2\), and \(y = 2x\). Evaluate the function on each line.- On \(x = 0\), the function becomes: \[ f(0, y) = y^2 - 4y + 1 \] To find critical points on this line, complete the square: \[ (y - 2)^2 - 3 = 0 \] Critical point at \(y = 2\) (vertex). Evaluate \(f(0, 0) = 1\) and \(f(0, 2) = -3\).- On \(y = 2\), the function becomes: \[ f(x, 2) = 2x^2 - 4x + 4 - 8 + 1 = 2x^2 - 4x - 3 \] Completing the square gives: \[ (x-1)^2 - 4 \] Critical point at \(x=1\). Evaluate \(f(0, 2) = -3\) and \(f(1, 2) = -3\).- On \(y = 2x\), the function becomes: \[ f(x, 2x) = 2x^2 - 4x + 4x^2 - 8x + 1 = 6x^2 - 12x + 1 \] Completing the square: \[ 6(x - 1)^2 - 5 \] Critical point at \(x=1\). Evaluate \(f(0, 0) = 1\) and \(f(1, 2) = -3\).
03

Compare Function Values

The evaluated points are:1. \(f(0, 0) = 1\)2. \(f(0, 2) = -3\)3. \(f(1, 2) = -3\)The maximum value is \(1\) at \((0, 0)\) and the minimum value is \(-3\) at both \((0, 2)\) and \((1, 2)\).

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Key Concepts

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

Critical Points
In calculus, critical points are where the derivative(s) of a function are zero or undefined. For multivariable functions, these are found using partial derivatives and set within a particular domain.
To locate the critical points of a function like our example, we compute the partial derivatives with respect to each variable. For our function, the partial derivative concerning \( x \) is \( 4x - 4 \). When set to zero, this gives us \( x = 1 \). Similarly, for \( y \), the partial derivative \( f_y = 2y - 4 \) leads us to \( y = 2 \).
This means our critical point is \( (1, 2) \). However, it is crucial to check if this point lies within the given boundary, which in our case, it does as it fits within the triangular region defined by the problem.
Partial Derivatives
Partial derivatives are a key concept in finding critical points of multivariable functions. They demonstrate how a function changes as each individual variable changes, holding others constant.
For the function \( f(x, y) = 2x^2 - 4x + y^2 - 4y + 1 \), the partial derivative with respect to \( x \) was calculated as \( 4x - 4 \). This tells us how the function changes as \( x \) changes, assuming \( y \) is constant. For \( y \), it was \( 2y - 4 \). Both these derivatives are instrumental in locating where the slope of the function is zero for each variable.
  • They form the basic tools to find stationary points that might correspond to peaks, troughs, or saddle points of the function.
  • A critical point is categorized as such only if both partial derivatives are zero.
Closed Triangular Region
The concept of a closed triangular region pertains to the specific area where we evaluate the function to find its absolute extrema. In this exercise, the boundaries are set by the lines \( x = 0 \), \( y = 2 \), and \( y = 2x \).
The closed nature of this region means we're considering both the interior and edges of the triangle. Thus, after determining critical points, the function needs to be evaluated on the boundaries as well.
This ensures that we don't miss any absolute maxima or minima that could occur along the edges of the triangle, not just within the interior. We assess the function on these lines by plugging into the function whatever simplifies and represents each constraint.

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Most popular questions from this chapter

Each of Exercises \(63-66\) gives a function \(f(x, y, z)\) and a positive number \(\epsilon .\) In each exercise, show that there exists a \(\delta>0\) such that for all \((x, y, z)\) , $$ \quad \sqrt{x^{2}+y^{2}+z^{2}}<\delta \Rightarrow|f(x, y, z)-f(0,0,0)|<\epsilon $$ $$ f(x, y, z)=\frac{x+y+z}{x^{2}+y^{2}+z^{2}+1}, \quad \epsilon=0.015 $$

Submarine sinkings The data in Table 14.4 show the results of a historical study of German submarines sunk by the U.S. Navy during 16 consecutive months of World War II. The data given for each month are the number of reported sinkings and the number of actual sinkings. The number of submarines sunk was slightly greater than the Navy's reports implied. Find a least squares line for estimating the number of actual sinkings from the number of reported sinkings.

Use Taylor's formula to find a quadratic approximation of \(f(x, y)=\cos x \cos y\) at the origin. Estimate the error in the approximation if \(|x| \leq 0.1\) and \(|y| \leq 0.1 .\)

In Exercises \(1-10\) , use Taylor's formula for \(f(x, y)\) at the origin to find quadratic and cubic approximations of \(f\) near the origin. $$ f(x, y)=\frac{1}{1-x-y+x y} $$

The Sandwich Theorem for functions of two variables states that if \(g(x, y) \leq f(x, y) \leq h(x, y)\) for all \((x, y) \neq\left(x_{0}, y_{0}\right)\) in a disk centered at \(\left(x_{0}, y_{0}\right)\) and if \(g\) and \(h\) have the same finite limit \(L\) as \((x, y) \rightarrow\left(x_{0}, y_{0}\right),\) then $$ \lim _{(x, y) \rightarrow\left(x_{0}, y_{0}\right)} f(x, y)=L $$ Use this result to support your answers to the questions in Exercises \(45-48 .\) Does knowing that $$ 1-\frac{x^{2} y^{2}}{3}<\frac{\tan ^{-1} x y}{x y}<1 $$ tell you anything about $$\lim _{(x, y) \rightarrow(0,0)} \frac{\tan ^{-1} x y}{x y} ?$$ Give reasons for your answer.
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