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Find the absolute maximum and minimum of the function \(f(x, y)=x^{2}+\) \(y^{2}\) subject to the constraint \(x^{4}+y^{4}=6561\). As usual, ignore unneeded answer blanks, and list points in lexicographic order. Absolute minimum value: ____________________. attained at ( ____________, ______________) ( ____________, ) Absolute minimum value: ____________________. attained at ( ____________, ______________) ( ____________, )

Short Answer

Expert verified
Absolute minimum value: 81. attained at (-9, 0), (0, -9) Absolute minimum value: 81. attained at (9, 0), (0, 9)

Step by step solution

01

Set up the Lagrange function

The Lagrange function is given by \(L(x, y, \lambda) = f(x, y) - \lambda g(x, y)\), where g(x, y) is the constraint function. In our case, f(x, y) = \(x^2 + y^2\) and g(x, y) = \(x^4 + y^4 - 6561\). So, the Lagrange function is: \(L(x, y, \lambda) = x^2 + y^2 - \lambda (x^4 + y^4 - 6561)\).
02

Compute the gradient of f(x, y) and the constraint function g(x, y)

We need to compute the gradient, which is the partial derivatives with respect to x, y, and \(\lambda\). So we have the following partial derivatives: \(\frac{\partial L}{\partial x} = 2x - \lambda 4x^3 = 0\) \(\frac{\partial L}{\partial y} = 2y - \lambda 4y^3 = 0\) \(\frac{\partial L}{\partial \lambda} = x^4 + y^4 - 6561 = 0\)
03

Equate the gradients and solve for x and y

From the partial derivatives with respect to x and y, we get two equations in x and y: \(2x - \lambda 4x^3 = 0 \Rightarrow x(1 - 2\lambda x^2) = 0\) \(2y - \lambda 4y^3 = 0 \Rightarrow y(1 - 2\lambda y^2) = 0\) Also, we have the constraint equation: \(x^4 + y^4 = 6561\).
04

Find the critical points (x, y)

Either x = 0 or \(1 - 2\lambda x^2 = 0\). Either y = 0 or \(1 - 2\lambda y^2 = 0\). Case 1: (x = 0, y ≠ 0) – In this case, when plugging x = 0 into constraint equazione, we get \(y^4 = 6561\). Taking the fourth root, we have y = ±9, x = 0. (0, -9) and (0, 9) are critical points. Case 2: (x ≠ 0, y = 0) – Similar to Case 1. The critical points are (-9, 0) and (9, 0). Case 3: Both x and y are non-zero. We have \(1 - 2\lambda x^2 = 0\) and \(1 - 2\lambda y^2 = 0\). We can solve these equations to find that \(x^2 = y^2\). But, we also have the constraint equation \(x^4 + y^4 = 6561\). Combining the two equations, we find that there are no solutions satisfying both equations. So the critical points are (-9, 0), (9, 0), (0, -9), and (0, 9).
05

Evaluate f(x, y) at the critical points

Now we evaluate the function \(f(x,y) = x^2 + y^2\) at these critical points: f(-9, 0) = (-9)^2 + 0 = 81 f(9, 0) = 9^2 + 0 = 81 f(0, -9) = 0 + (-9)^2 = 81 f(0, 9) = 0 + 9^2 = 81 Since all the values of the function at the critical points are equal, we have: Absolute minimum value: 81, attained at (-9, 0) and (0, -9) Absolute maximum value: 81, attained at (9, 0) and (0, 9)

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Key Concepts

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

Multivariable Calculus
Multivariable calculus is an extension of calculus that deals with functions of multiple variables, such as two or more, rather than just one. In this context, we're considering the function \( f(x, y) = x^2 + y^2 \). By understanding how this function behaves as both \(x\) and \(y\) change, we can explore a range of features like slopes, rates of change, and extrema (or extreme values).
When dealing with multivariable functions, one key element is understanding the concept of partial derivatives. These derivatives help analyze how a function changes with respect to one variable at a time, keeping the others constant. Here, in our exercise, when we compute gradients, we're essentially looking at these partial derivatives.
This foundational knowledge of multivariable calculus is crucial for tackling more complex topics such as optimization and constraints, as it provides the necessary tools to understand changes in quantity across multiple dimensions.
Optimization
Optimization is all about finding the maximum or minimum values of a function. In other words, it's like hunting for the 'best' or 'worst' possible outcomes that a function can give, based on certain conditions.
There are two main types of optimization: unconstrained and constrained. Unconstrained optimization looks for maxima or minima without any restrictions. Constrained optimization, like in our problem here, finds these extremes but within given limitations.
In the problem we've got, we're asked to find the extreme values of a function \( f(x, y) = x^2 + y^2 \). But we're bound by a constraint: \( x^4 + y^4 = 6561 \). This means, our task is not just to find the maxima and minima but to do so while adhering to this specific condition.
That's where the method of Lagrange multipliers comes into play. It's a nifty tool that helps in dealing with such constraints in optimization problems, ensuring that we don't step outside the bounds of the constraints set for us.
Constraint
Constraints are the limits or conditions placed upon the values that variables in a function can assume, defining the region of interest for our solutions. In the exercise we're tackling, the constraint is given by \( x^4 + y^4 = 6561 \).
Such constraints are crucial because they restrict the domain and guide where the function's extreme values are to be found. Constraints can be in different forms, such as equations like we see here, or inequalities which are quite common in other problems.
These constraints play a role not only in practical applications, like maximizing profit or minimizing cost in a real-world scenario, but they are also essential in ensuring that the solutions to the mathematical problems are realistic and applicable. In mathematical optimization, respecting constraints is necessary to find valid solutions that comply with any given conditions or limitations.
Critical Points
Critical points are specific points on a function where the first derivative (in each variable) is zero or undefined. In multivariable calculus, these are the points we scrutinize to determine potential extreme values—either maximums, minimums, or saddle points.
For the given function \( f(x, y) = x^2 + y^2 \) and the constraint \( x^4 + y^4 = 6561 \), we seek critical points by setting up the Lagrange function and finding where the gradients of our function and constraint are parallel.
Here, we calculated the critical points as \((-9, 0)\), \((9, 0)\), \((0, -9)\), and \((0, 9)\). After examining the function values at these points, we determined all yield the same value, meaning the extremes coincide at each of these points.
Finding critical points is a way to zero in on where changes happen in the function and where those bound by conditions give the desired extremes. This involves analyzing the nature of these points and verifying if they are indeed extremal (highest or lowest) given the context.

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Most popular questions from this chapter

The Heat Index, \(I,\) (measured in apparent degrees \(F)\) is a function of the actual temperature \(T\) outside (in degrees \(\mathrm{F}\) ) and the relative humidity \(H\) (measured as a percentage). A portion of the table which gives values for this function, \(I=I(T, H),\) is reproduced in Table \(10.2 .10 .\) $$\begin{array}{ccccc}\hline T \downarrow \backslash H \rightarrow & 70 & 75 & 80 & 85 \\ \hline 90 & 106 & 109 & 112 & 115 \\\\\hline 92 & 112 & 115 & 119 & 123 \\\\\hline 94 & 118 & 122 & 127 & 132 \\\\\hline 96 & 125 & 130 & 135 & 141 \\\\\hline\end{array}$$ a. State the limit definition of the value \(I_{T}(94,75)\). Then, estimate \(I_{T}(94,75),\) and write one complete sentence that carefully explains the meaning of this value, including its units. b. State the limit definition of the value \(I_{H}(94,75)\). Then, estimate \(I_{H}(94,75),\) and write one complete sentence that carefully explains the meaning of this value, including its units. c. Suppose you are given that \(I_{T}(92,80)=3.75\) and \(I_{H}(92,80)=0.8\). Estimate the values of \(I(91,80)\) and \(I(92,78)\). Explain how the partial derivatives are relevant to your thinking. d. On a certain day, at 1 p.m. the temperature is 92 degrees and the relative humidity is \(85 \%\). At 3 p.m., the temperature is 96 degrees and the relative humidity \(75 \% .\) What is the average rate of change of the heat index over this time period, and what are the units on your answer? Write a sentence to explain your thinking.

In the following questions, we determine and apply the linearization for several different functions. a. Find the linearization \(L(x, y)\) for the function \(f\) defined by \(f(x, y)=\) \(\cos (x)\left(2 e^{2 y}+e^{-2 y}\right)\) at the point \(\left(x_{0}, y_{0}\right)=(0,0) .\) Use the linearization to estimate the value of \(f(0.1,0.2)\). Compare your estimate to the actual value of \(f(0.1,0.2)\) b. The Heat Index, \(I,\) (measured in apparent degrees \(\mathrm{F}\) ) is a function of the actual temperature \(T\) outside (in degrees \(\mathrm{F}\) ) and the relative humidity \(H\) (measured as a percentage). A portion of the table which gives values for this function, \(I=I(T, H),\) is provided in Table 10.4 .13 $$\begin{array}{ccccc}\hline T \downarrow \backslash H \rightarrow & 70 & 75 & 80 & 85 \\ \hline 90 & 106 & 109 & 112 & 115 \\\\\hline 92 & 112 & 115 & 119 & 123 \\ \hline 94 & 118 & 122 & 127 & 132 \\\\\hline 96 & 125 & 130 & 135 & 141 \\\\\hline\end{array}$$ Suppose you are given that \(I_{T}(94,75)=3.75\) and \(I_{H}(94,75)=0.9\). Use this given information and one other value from the table to estimate the value of \(I(93.1,77)\) using the linearization at (94,75) . Using proper terminology and notation, explain your work and thinking. c. Just as we can find a local linearization for a differentiable function of two variables, we can do so for functions of three or more variables. By extending the concept of the local linearization from two to three variables, find the linearization of the function \(h(x, y, z)=e^{2 x}(y+\) \(z^{2}\) ) at the point \(\left(x_{0}, y_{0}, z_{0}\right)=(0,1,-2) .\) Then, use the linearization to estimate the value of \(h(-0.1,0.9,-1.8)\).

Consider the function $$f(x, y)=\left(18 x-x^{2}\right)\left(18 y-y^{2}\right)$$ Find and classify all critical points of the function. If there are more blanks than critical points, leave the remaining entries blank. \(f_{x}=\) _________. \(f_{y}=\) _________. \(f_{x x}=\) _________. \(f_{x y}=\) _________. \(f_{y y}=\) _________. There are several critical points to be listed. List them lexicograhically, that is in ascending order by x-coordinates, and for equal x-coordinates in ascending order by y-coordinates (e.g., (1,1),(2,-1),(2,3) is a correct order) The critical point with the smallest x-coordinate is (_____________, ____________) Classification: (_____________, ____________) (local minimum, local maximum, saddle point, cannot be determined) The critical point with the next smallest x-coordinate is (_____________, ____________) Classification: (_____________, ____________) (local minimum, local maximum, saddle point, cannot be determined) The critical point with the next smallest \(\mathrm{x}\) -coordinate is (_____________, ____________) Classification: (_____________, ____________) (local minimum, local maximum, saddle point, cannot be determined) The critical point with the next smallest x-coordinate is (_____________, ____________) Classification: (_____________, ____________) (local minimum, local maximum, saddle point, cannot be determined) The critical point with the next smallest \(x\) -coordinate is (_____________, ____________) Classification: (_____________, ____________) (local minimum, local maximum, saddle point, cannot be determined)

In this section we argued that if \(f=f(x, y)\) is a function of two variables and if \(f_{x}\) and \(f_{y}\) both exist and are continuous in an open disk containing the point \(\left(x_{0}, y_{0}\right),\) then \(f\) is differentiable at \(\left(x_{0}, y_{0}\right) .\) This condition ensures that \(f\) is differentiable at \(\left(x_{0}, y_{0}\right),\) but it does not define what it means for \(f\) to be differentiable at \(\left(x_{0}, y_{0}\right) .\) In this exercise we explore the definition of differentiability of a function of two variables in more detail. Throughout, let \(g\) be the function defined by \(g(x, y)=\sqrt{|x y|}\) a. Use appropriate technology to plot the graph of \(g\) on the domain \([-1,1] \times[-1,1] .\) Explain why \(g\) is not locally linear at (0,0) b. Show that both \(g_{x}(0,0)\) and \(g_{y}(0,0)\) exist. If \(g\) is locally linear at \((0,0),\) what must be the equation of the tangent plane \(L\) to \(g\) at (0,0)\(?\) c. Recall that if a function \(f=f(x)\) of a single variable is differentiable at \(x=x_{0},\) then $$f^{\prime}\left(x_{0}\right)=\lim _{h \rightarrow 0} \frac{f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h}$$ exists. We saw in single variable calculus that the existence of \(f^{\prime}\left(x_{0}\right)\) means that the graph of \(f\) is locally linear at \(x=x_{0}\). In other words, the graph of \(f\) looks like its linearization \(L(x)=f\left(x_{0}\right)+\) \(f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)\) for \(x\) close to \(x_{0} .\) That is, the values of \(f(x)\) can be closely approximated by \(L(x)\) as long as \(x\) is close to \(x_{0}\). We can measure how good the approximation of \(L(x)\) is to \(f(x)\) with the error function $$E(x)=L(x)-f(x)=f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)-f(x)$$ As \(x\) approaches \(x_{0}, E(x)\) approaches \(f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)(0)-f\left(x_{0}\right)=0\), and so \(L(x)\) provides increasingly better approximations to \(f(x)\) as \(x\) gets closer to \(x_{0} .\) Show that, even though \(g(x, y)=\sqrt{|x y|}\) is not locally linear at \((0,0),\) its error term $$ E(x, y)=L(x, y)-g(x, y) $$ at (0,0) has a limit of 0 as \((x, y)\) approaches \((0,0) .\) (Use the linearization you found in part (b).) This shows that just because an error term goes to 0 as \((x, y)\) approaches \(\left(x_{0}, y_{0}\right),\) we cannot conclude that a function is locally linear at \(\left(x_{0}, y_{0}\right)\). d. As the previous part illustrates, having the error term go to 0 does not ensure that a function of two variables is locally linear. Instead, we need a notation of a relative error. To see how this works, let us return to the single variable case for a moment and consider \(f=f(x)\) as a function of one variable. If we let \(x=x_{0}+h,\) where \(|h|\) is the distance from \(x\) to \(x_{0}\), then the relative error in approximating \(f\left(x_{0}+h\right)\) with \(L\left(x_{0}+h\right)\) is $$\frac{E\left(x_{0}+h\right)}{h}$$ Show that, for a function \(f=f(x)\) of a single variable, the limit of the relative error is 0 as \(h\) approaches 0 . e. Even though the error term for a function of two variables might have a limit of 0 at a point, our example shows that the function may not be locally linear at that point. So we use the concept of relative error to define differentiability of a function of two variables. When we consider differentiability of a function \(f=f(x, y)\) at a point \(\left(x_{0}, y_{0}\right),\) then if \(x=x_{0}+h\) and \(y=y_{0}+k,\) the distance from \((x, y)\) to \(\left(x_{0}, y_{0}\right)\) is \(\sqrt{h^{2}+k^{2}}\)

In a simple electric circuit, Ohm's law states that \(V=I R,\) where \(\mathrm{V}\) is the voltage in volts, I is the current in amperes, and \(\mathrm{R}\) is the resistance in ohms. Assume that, as the battery wears out, the voltage decreases at 0.03 volts per second and, as the resistor heats up, the resistance is increasing at 0.02 ohms per second. When the resistance is 100 ohms and the current is 0.02 amperes, at what rate is the current changing? ___________ amperes per second .

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