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Chapter 15: Problem 70

Find the dimensions of the rectangular box with maximum volume in the first octant with one vertex at the origin and the opposite vertex on the ellipsoid \(36 x^{2}+4 y^{2}+9 z^{2}=36\)

Short Answer

Expert verified
Based on the given ellipsoid constraint and the method of Lagrange multipliers, we found that a rectangular box located in the first octant will have maximum volume if its dimensions are approximately \(x=\sqrt{\frac{1}{163}}\), \(y=9\sqrt{\frac{1}{163}}\), and \(z=18\sqrt{\frac{1}{163}}\).

Step by step solution

01

Understanding the ellipsoid constraint

Our constraint is an ellipsoid defined by \(36 x^{2} + 4 y^{2} + 9 z^{2} = 36\). Dividing each term by \(36\), we get: \((x^{2} + \frac{y^2}{9} + \frac{z^2}{4}) = 1\). This is an ellipsoid centered at the origin with semi-axes of lengths \(a=1\), \(b=3\), and \(c=2\) in the \(x\), \(y\), and \(z\) directions, respectively.
02

Volume function for the rectangular box

Let \((x, y, z)\) be the coordinates of the opposite vertex of the rectangular box on the ellipsoid. Then, the volume \(V\) of the rectangular box can be represented by the product of its side lengths, which are \(x\), \(y\), and \(z\). So, we have \(V(x, y, z) = xyz\).
03

Applying Lagrange multipliers

We wish to maximize the volume \(V(x, y, z)\) subject to the constraint \((x^{2} + \frac{y^2}{9} + \frac{z^2}{4}) = 1\). To do so, we'll use the method of Lagrange multipliers. Define the Lagrangian function \(L(x, y, z, \lambda) = xyz - \lambda (x^2 + \frac{y^2}{9} + \frac{z^2}{4} - 1)\). To find the maximum volume, we need to solve the system of equations given by the gradient of \(L\) being equal to the zero vector: 1. \(\frac{\partial L}{\partial x} = yz - 2\lambda x = 0\) 2. \(\frac{\partial L}{\partial y} = xz - \frac{2\lambda y}{9} = 0\) 3. \(\frac{\partial L}{\partial z} = xy - \frac{\lambda z}{2} = 0\) 4. Constraint equation: \(x^2 + \frac{y^2}{9} + \frac{z^2}{4} = 1\)
04

Solving the system of equations

First, note that none of \(x\), \(y\), or \(z\) can be zero, as this would result in a zero-volume box. Divide equation 1 by \(x\), equation 2 by \(y\), and equation 3 by \(z\): 1'. \(yz - 2\lambda = 0\) 2'. \(xz - \frac{2\lambda}{9} = 0\) 3'. \(xy - \frac{\lambda}{2} = 0\) Now, divide equation 1' by equation 2' and equation 3' by equation 1': 1''. \(\frac{y}{x} - 9 = 0\) 2''. \(\frac{z}{y} - 2 = 0\) Solve for \(y\) and \(z\) in terms of \(x\): 1'''. \(y = 9x\) 2'''. \(z = 2y = 18x\) Now, substitute these expressions for \(y\) and \(z\) in the constraint equation: \(x^2 + \frac{(9x)^2}{9} + \frac{(18x)^2}{4} = 1\) \(x^2 + 81x^2 + 81x^2 = 1\) Combine like terms and solve for \(x\): \(163x^2 = 1\) \(x = \sqrt{\frac{1}{163}}\) Now, compute \(y\) and \(z\): \(y = 9x = 9\sqrt{\frac{1}{163}}\) \(z = 18x = 18\sqrt{\frac{1}{163}}\)
05

Dimensions of the rectangular box

The dimensions of the rectangular box with maximum volume in the first octant with one vertex at the origin and the opposite vertex on the ellipsoid \(36 x^{2} + 4 y^{2} + 9 z^{2} = 36\) are \(x=\sqrt{\frac{1}{163}}\), \(y=9\sqrt{\frac{1}{163}}\), and \(z=18\sqrt{\frac{1}{163}}\).

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Key Concepts

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

Ellipsoid Constraint
An ellipsoid is a three-dimensional surface, often thought of as a stretched or squashed sphere. It can be defined by the equation \(Ax^2 + By^2 + Cz^2 = D\). In the context of our problem, the ellipsoid is given by the equation \(36 x^{2} + 4 y^{2} + 9 z^{2} = 36\).
This equation represents an ellipsoid centered at the origin where the values of \(A\), \(B\), and \(C\) determine the shape. By simplifying this, we divide the whole equation by 36, yielding \(x^2 + \frac{y^2}{9} + \frac{z^2}{4} = 1\).
  • The coefficients dictate the lengths of the semi-axes along the respective axes.
  • Here, \(x\), \(y\), and \(z\) are scaled by factors of 1, 3, and 2 respectively.
  • This gives us a better understanding of how the ellipsoid is oriented and how its constraint comes into play.
Recognizing such constraints is essential for determining how figures like boxes fit within these boundaries, particularly in optimization problems.
Volume Maximization
Volume maximization deals with finding the largest possible volume a three-dimensional object can have under given constraints. For our problem, we're looking to maximize the volume of a rectangular box within the bounds set by an ellipsoid.
The volume \(V\) of a rectangular box with side lengths \(x\), \(y\), and \(z\) is calculated as \(V = xyz\).
  • The challenge here is to stretch the dimensions of the box as much as possible.
  • However, it must still conform to the shape defined by the ellipsoid (our constraint).
  • This requires a method to balance the dimensions so as to not violate the ellipsoid's edge.

By utilizing methods like Lagrange multipliers, we can systematically find such a balance. This process ensures we can find dimensions that yield the maximum volume possible under the given constraint.
Gradient
The gradient is a critical concept in optimization and calculus, representing the multidimensional rate of change of a function. In our case, the function we want to maximize is the box's volume, while acknowledging the imposed ellipsoid constraint. When using Lagrange multipliers, we leverage gradients to find optimal solutions.
The Lagrangian \(L(x, y, z, \lambda)\) is built with both the volume function and the constraint, \(L = xyz - \lambda(x^2 + \frac{y^2}{9} + \frac{z^2}{4} - 1)\).
  • To maximize, we need all partial derivatives of \(L\) to be zero 鈥� this forms a system of equations.
  • The gradient \(abla L\) becomes zero at the points where there is no change in \(L\), indicating potential maxima or minima.
  • These points need to satisfy both the volume function and the constraint, allowing us to determine optimal dimensions.

Understanding gradients enables us to navigate these changes and find where the function's value is maximized, given constraints like the ellipsoid. This powerful tool allows one to systematically explore function behavior across dimensions.

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Most popular questions from this chapter

A baseball pitcher's earned run average (ERA) is \(A(e, i)=9 e / i\), where \(e\) is the number of earned runs given up by the pitcher and \(i\) is the number of innings pitched. Good pitchers have low ERAs. Assume \(e \geq 0\) and \(i>0\) are real numbers. a. The single-season major league record for the lowest ERA was set by Dutch Leonard of the Detroit Tigers in \(1914 .\) During that season, Dutch pitched a total of 224 innings and gave up just 24 earned runs. What was his ERA? b. Determine the ERA of a relief pitcher who gives up 4 earned runs in one- third of an inning. c. Graph the level curve \(A(e, i)=3\) and describe the relationship between \(e\) and \(i\) in this case.

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Probability of at least one encounter Suppose in a large group of people, a fraction \(0 \leq r \leq 1\) of the people have flu. The probability that in \(n\) random encounters you will meet at least one person with flu is \(P=f(n, r)=1-(1-r)^{n} .\) Although \(n\) is a positive integer, regard it as a positive real number. a. Compute \(f_{r}\) and \(f_{n^{*}}\) b. How sensitive is the probability \(P\) to the flu rate \(r ?\) Suppose you meet \(n=20\) people. Approximately how much does the probability \(P\) increase if the flu rate increases from \(r=0.1\) to \(r=0.11(\text { with } n \text { fixed }) ?\) c. Approximately how much does the probability \(P\) increase if the flu rate increases from \(r=0.9\) to \(r=0.91\) with \(n=20 ?\) d. Interpret the results of parts (b) and (c).
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