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Chapter 13: Problem 34

Applications of Lagrange multipliers Use Lagrange multipliers in the following problems. When the domain of the objective function is unbounded or open, explain why you have found an absolute maximum or minimum value. Maximum volume cylinder in a sphere Find the dimensions of a right circular cylinder of maximum volume that can be inscribed in a sphere of radius 16.

Short Answer

Expert verified
Answer: The dimensions of the inscribed right circular cylinder with maximum volume are a radius of 32/3 units and a height of 32/3 units.

Step by step solution

01

1. Define the objective function and constraint

Let the radius of the cylinder be r and its height be h. Our objective is to maximize its volume, V. Therefore, the objective function is: V(r, h) = 蟺r^2h Since the cylinder is inscribed in a sphere with radius 16, that means the sum of the cylinder's radius and half of its height must equal the sphere's radius. This is our constraint: g(r, h) = r + h/2 = 16
02

2. Introduce the Lagrange multiplier

Define the Lagrangian function with the multiplier 位: L(r, h, 位) = 蟺r^2h - 位(r + h/2 - 16)
03

3. Find the partial derivatives of L and set them to zero

We need to find the partial derivatives of L with respect to r, h, and 位 and set them to zero: 鈭侺/鈭俽 = 2蟺rh - 位 = 0 鈭侺/鈭俬 = 蟺r^2 - 位/2 = 0 鈭侺/鈭偽� = r + h/2 - 16 = 0
04

4. Solve the system of equations

We have the following system of equations: (1) 2蟺rh - 位 = 0 (2) 蟺r^2 - 位/2 = 0 (3) r + h/2 - 16 = 0 Now, from (3) we can express h using r: h = 32 - 2r Substitute this expression of h in (1): 2蟺r(32 - 2r) - 位 = 0 Now, solving (2) for 位 and substituting it back into the equation above: 位 = 2蟺r^2 2蟺r(32 - 2r) - 2蟺r^2 = 0 Divide by 2蟺r: 32 - 2r - r = 0 Solve for r: r = 32/3 Now, substitute the value of r back into the h expression: h = 32 - 2(32/3) h = 32/3
05

5. Conclude the solution

The dimensions of the inscribed right circular cylinder with maximum volume in a sphere of radius 16 are as follows: - Radius: r = 32/3 units - Height: h = 32/3 units This cylinder has a maximum volume of V = 蟺(32/3)^2(32/3) cubic units.

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Key Concepts

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

Objective Function
An objective function is a key element in optimization problems and is what you're trying to maximize or minimize. In this exercise, our goal is to maximize the volume of a right circular cylinder. The formula for the volume, \( V \), of a cylinder is expressed as \( V(r, h) = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height of the cylinder.
This formula represents our objective function. The challenge is to find values for \( r \) and \( h \) that provide the highest possible volume while satisfying certain conditions.
Constraint Optimization
Constraint optimization deals with finding the best solution to a problem within a set of restrictions or limits. In our problem, we are working with a cylinder inscribed in a sphere, meaning the dimensions of the cylinder are constrained by the size of the sphere.
For this exercise, the constraint is derived from the geometric condition that the cylinder must fit inside the sphere. The sphere has a radius of 16, and the constraint equation is \( g(r, h) = r + \frac{h}{2} = 16 \). This equation ensures that the cylinder fits within the sphere's boundaries by relating the radius and height accordingly.
  • Constraint optimization allows us to explore possibilities that adhere to real-world limitations.
  • It helps to determine feasible solutions that maximize or minimize the objective function.
Analyzing these constraints alongside the objective function is crucial to finding the optimal solution.
Cylinder Inscribed in Sphere
A cylinder inscribed in a sphere is a classic geometry problem involving fitting a three-dimensional figure within another. To solve such problems, it's essential to know about the relations among the radius and height of the cylinder and the radius of the sphere.
A sphere's radius sets a limit on how large the cylinder can be while still fitting inside. The condition \( r + \frac{h}{2} = 16 \) expresses this fitting constraint mathematically. By inscribing a cylinder within a sphere, we ensure that:
  • The height and radius of the cylinder are adjustable but constrained by the sphere.
  • Any improvement in the dimensions should respect this geometric condition.
In optimization, such constraints aid in determining plausible dimensions and achieving the goal, which is to maximize the cylinder's volume.

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Let \(w=f(x, y, z)=2 x+3 y+4 z\), which is defined for all \((x, y, z)\) in \(\mathbb{R}^{3}\). Suppose that we are interested in the partial derivative \(w_{x}\) on a subset of \(\mathbb{R}^{3}\), such as the plane \(P\) given by \(z=4 x-2 y .\) The point to be made is that the result is not unique unless we specify which variables are considered independent. a. We could proceed as follows. On the plane \(P\), consider \(x\) and \(y\) as the independent variables, which means \(z\) depends on \(x\) and \(y,\) so we write \(w=f(x, y, z(x, y)) .\) Differentiate with respect to \(x\) holding \(y\) fixed to show that \(\left(\frac{\partial w}{\partial x}\right)_{y}=18,\) where the subscript \(y\) indicates that \(y\) is held fixed. b. Alternatively, on the plane \(P,\) we could consider \(x\) and \(z\) as the independent variables, which means \(y\) depends on \(x\) and \(z,\) so we write \(w=f(x, y(x, z), z)\) and differentiate with respect to \(x\) holding \(z\) fixed. Show that \(\left(\frac{\partial w}{\partial x}\right)_{z}=8,\) where the subscript \(z\) indicates that \(z\) is held fixed. c. Make a sketch of the plane \(z=4 x-2 y\) and interpret the results of parts (a) and (b) geometrically. d. Repeat the arguments of parts (a) and (b) to find \(\left(\frac{\partial w}{\partial y}\right)_{x}\), \(\left(\frac{\partial w}{\partial y}\right)_{z},\left(\frac{\partial w}{\partial z}\right)_{x},\) and \(\left(\frac{\partial w}{\partial z}\right)_{y}\).
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