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Chapter 10: Problem 4

Find the absolute maximum and minimum of the function \(f(x, y)=x^{2}-\) \(y^{2}\) subject to the constraint \(x^{2}+y^{2}=361 .\) 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: \(-361\) attained at \((0, -19)\) and \((0, 19)\) Absolute maximum value: \(361\) attained at \((-19, 0)\) and \((19, 0)\)

Step by step solution

01

Compute the gradients of f and g

Find the gradient of the function \(f(x, y) = x^2 - y^2\) and the constraint \(g(x, y) = x^2 + y^2 - 361\): Gradient of f: \(\nabla f(x, y) = \left( \frac{\partial f}{\partial x},\frac{\partial f}{\partial y} \right) = (2x, -2y)\) Gradient of g: \(\nabla g(x, y) = \left( \frac{\partial g}{\partial x},\frac{\partial g}{\partial y} \right) = (2x, 2y)\)
02

Apply the method of Lagrange Multipliers

To solve the system of equations using the Lagrange Multipliers method, we need to equate the gradient of f to a scalar multiple of the gradient of g and satisfy the constraint: \(\nabla f(x, y) = \lambda \nabla g(x, y)\), where \(\lambda\) is a constant (Lagrange multiplier). From the gradients, we have: \(2x = \lambda (2x)\) \(-2y = \lambda (2y)\) \(x^2 + y^2 = 361\)
03

Solve the system of equations

Solve the system of equations: If \(\lambda = 1\), from the first equation, we get \(2x = 2x\), which always holds. From the second equation, we get \(-2y = 2y\), which implies \(y = 0\). Plugging \(y = 0\) to the constraint, we get \(x^2 = 361\), so \(x = \pm{19}\). Thus, we have two points: \((19, 0)\) and \((-19, 0)\). If \(\lambda = -1\), from the first equation, we get \(2x = -2x\), which implies \(x = 0\). From the second equation, we get \(-2y = -2y\), which always holds. Plugging \(x = 0\) to the constraint, we get \(y^2 = 361\), so \(y = \pm{19}\). Thus, we have two points: \((0, 19)\) and \((0, -19)\).
04

Substitute points into f(x, y) and find the maximum and minimum values

Substitute the points \((19, 0), (-19, 0), (0, 19), (0, -19)\) into the function \(f(x, y) = x^2 - y^2\): f(19, 0) = \((19)^2 - (0)^2 = 361\) f(-19, 0) = \((-19)^2 - (0)^2 = 361\) f(0, 19) = \((0)^2 - (19)^2 = -361\) f(0, -19) = \((0)^2 - (-19)^2 = -361\) From the above calculations, we have: Absolute minimum value: \(-361\) attained at \((0, -19)\) and \((0, 19)\) Absolute maximum value: \(361\) attained at \((-19, 0)\) and \((19, 0)\)

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Key Concepts

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

Gradient
The gradient is a fundamental concept in multivariable calculus that describes the directional rate of change of a function. When dealing with optimization problems and constraints, the gradient plays a crucial role. For a function of two variables, like our function here, the gradient \( abla f(x, y) \) is a vector consisting of the partial derivatives of the function with respect to each variable.

In our exercise, we have the function \( f(x, y) = x^2 - y^2 \). The gradient of \( f \) is \( abla f(x, y) = (2x, -2y) \). This vector points in the direction of the steepest ascent of the function. Similarly, for the constraint \( g(x, y) = x^2 + y^2 - 361 \), the gradient \( abla g(x, y) = (2x, 2y) \) was computed.

By comparing these gradients, we recognize the key insight of the Lagrange Multiplier method: the gradient of the function must be parallel to the gradient of the constraint at optimal points. This helps us find where maximum and minimum values occur within the constraints.
Constrained Optimization
Constrained optimization is a process used to find the maximum or minimum of a function subject to specific restrictions or constraints. In this exercise, we want to maximize and minimize the function \( f(x, y) = x^2 - y^2 \) under the constraint \( x^2 + y^2 = 361 \), which describes a circle with radius 19.

To solve this, we use Lagrange Multipliers. The idea is to set the gradient of our function equal to a constant, known as the Lagrange multiplier \( \lambda \), times the gradient of the constraint, \( abla g(x, y) \). This gives us the system of equations:
  • \( 2x = \lambda (2x) \)
  • \( -2y = \lambda (2y) \)
Additionally, we have the original constraint \( x^2 + y^2 = 361 \). By solving these equations, we find the specific points where the function reaches its extreme values while respecting the given constraint.
Critical Points
Critical points are the values of \( x \) and \( y \) that satisfy the conditions of the optimization problem and are candidates to be points of maximum or minimum. In our example, these are the points where the gradients are parallel, as determined by Lagrange's condition \( abla f(x, y) = \lambda abla g(x, y) \).

Through solving our system, we found two critical scenarios based on the values of \( \lambda \):
  • For \( \lambda = 1 \), we find points \((19, 0)\) and \((-19, 0)\)
  • For \( \lambda = -1 \), we find points \((0, 19)\) and \((0, -19)\)
These are the critical points on the contour of the constraint.

Substituting these points back into the function \( f(x, y) = x^2 - y^2 \), we calculate that \((19, 0)\) and \((-19, 0)\) yield the maximum value of 361, while \((0, 19)\) and \((0, -19)\) yield the minimum value of -361.

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Most popular questions from this chapter

For each value of \(\lambda\) the function \(h(x, y)=x^{2}+y^{2}-\lambda(2 x+8 y-20)\) has a minimum value \(m(\lambda)\). (a) Find \(m(\lambda)\) \(m(\lambda)=\) ___________. (Use the letter \(\boldsymbol{L}\) for \(\lambda\) in your expression. \()\) (b) For which value of \(\lambda\) is \(m(\lambda)\) the largest, and what is that maximum value? \(\lambda=\) \(\operatorname{maximum} m(\lambda)=\) _________. (c) Find the minimum value of \(f(x, y)=x^{2}+y^{2}\) subject to the constraint \(2 x+8 y=20\) using the method of Lagrange multipliers and evaluate \(\lambda\) \(\operatorname{minimum} f=\) ___________ \(\lambda=\) ___________

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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 \(h(-0.1,0.9,-1.8)\).

Use the chain rule to find \(\frac{\partial z}{\partial s}\) and \(\frac{\partial z}{\partial t},\) where $$z=e^{x y} \tan y, x=4 s+5 t, y=\frac{3 s}{3 t}$$ First the pieces: \(\frac{\partial z}{\partial x}=\) ________ \(\frac{\partial z}{\partial y}=\) \(\frac{\partial x}{\partial s}=\) ________ \(\frac{\partial x}{\partial t}=\) \(\frac{\partial y}{\partial s}=\) ________ \(\frac{\partial y}{\partial t}=\)

Suppose \(w=\frac{x}{y}+\frac{y}{z},\) where $$x=e^{5 t}, y=2+\sin (3 t), \text { and } z=2+\cos (6 t)$$ A) Use the chain rule to find \(\frac{d w}{d t}\) as a function of \(x, y, z,\) and \(t .\) Do not rewrite \(x, y,\) and \(z\) in terms of \(t,\) and do not rewrite \(e^{5 t}\) as \(x\). \(\frac{d w}{d t}=\) ____________. B) Use part A to evaluate \(\frac{d w}{\|}\) when \(t=0\).
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