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Chapter 14: Problem 58

Among all closed rectangular boxes of volume \(27 \mathrm{cm}^{3},\) what is the smallest surface area?

Short Answer

Expert verified
The smallest surface area is 54 cm虏, for a cube with sides of 3 cm.

Step by step solution

01

Understand the Problem

We need to find the dimensions of a rectangular box with volume 27 cm鲁 that minimize surface area. It's a problem involving calculus to optimize a variable.
02

Establish Variables

Let the dimensions of the box be \(x\), \(y\), and \(z\). Thus, the volume \(xyz = 27\). The surface area \(S\) is expressed as \(S = 2(xy + xz + yz)\).
03

Express Constraints

From the volume constraint, solve for one variable in terms of the others: \(z = \frac{27}{xy}\). This allows us to substitute back into the surface area formula.
04

Substitute and Simplify

Substitute \(z\) from the constraint into the surface area formula: \(S = 2(xy + x\frac{27}{xy} + y\frac{27}{xy}) = 2(xy + \frac{27}{x} + \frac{27}{y})\).
05

Apply Calculus

To minimize \(S\), calculate partial derivatives \(\frac{\partial S}{\partial x}\) and \(\frac{\partial S}{\partial y}\), set them to zero, and solve simultaneously to find critical points.
06

Solve for Critical Points

Set \(\frac{\partial S}{\partial x} = 2(y - \frac{27}{x^2} + \frac{27}{y}) = 0\) and \(\frac{\partial S}{\partial y} = 2(x - \frac{27}{y^2} + \frac{27}{x}) = 0\). Solve these equations for values of \(x\) and \(y\).
07

Find Dimensions

Through solving, it will be found that the optimal dimensions \(x = y = z = \sqrt[3]{27} = 3\), which means the box is a cube.
08

Verify with Surface Area Calculation

Compute the surface area using these dimensions: \(S = 2(3 \times 3 + 3 \times 3 + 3 \times 3) = 54 \mathrm{cm}^2)\). Check that this is the minimum surface area possible based on the given volume constraint.

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Key Concepts

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

Rectangular boxes
When diving into the world of geometry, one common shape you'll encounter is the rectangular box, often referred to as a cuboid. This shape features six rectangular faces meeting at 90-degree angles, creating a 3-dimensional object. In most problems like the one described, you'll deal with three critical dimensions: length (\(x\)), width (\(y\)), and height (\(z\)). Understanding these dimensions helps in establishing relationships and constraints, particularly for problems involving volume and surface area.
  • The volume of a rectangular box is given by \(Volume = x \, y \, z\), representing the space it occupies.
  • The surface area, on the other hand, represents the total area covering the outside of the box, calculated by \(Surface \ Area = 2(xy + xz + yz)\).
In optimization problems, these formulas serve as foundational equations that guide the process of finding the optimal dimensions according to the problem constraints.
Surface area minimization
Minimizing the surface area of a rectangular box while keeping a constant volume is a common optimization challenge in geometry and calculus. Understanding this concept involves manipulating the fundamental equations and applying constraints.
  • The main idea in minimization is to reduce the surface area formula \(S = 2(xy + xz + yz)\) as much as possible.
  • However, the constraint \(xyz = 27\) specifies that the box must have a constant volume.
To minimize surface area, we often express one variable in terms of others using the constraint. This substitution reduces the problem to functions of fewer variables, making calculus tools more applicable. By reducing these variables, it simplifies finding points at which the surface area is minimized.
Calculus applications
Calculus is pivotal in solving optimization problems, offering tools like derivatives to identify minimum or maximum values. In the context of minimizing the surface area of a rectangular box, calculus aids in fine-tuning dimensions that meet specific criteria.
  • Firstly, express the surface area \(S\) in terms of fewer variables using substitution derived from the volume constraint.
  • Then, take partial derivatives of \(S\) with respect to the chosen variables.
  • Set these derivatives equal to zero to find critical points鈥攑otential solutions where the minimum surface area could occur.
Solving the resulting equations typically leads you to find optimal dimensions. For this specific problem, it turns out that the dimensions that minimize the surface area are where each side of the box is equal, forming a cube. This solution not only satisfies the given constraint of the volume, but also achieves the smallest possible surface area.

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Most popular questions from this chapter

In Exercises \(57-60,\) use the limit definition of partial derivative to compute the partial derivatives of the functions at the specified points. $$ f(x, y)=\left\\{\begin{array}{ll}{\frac{\sin \left(x^{3}+y^{4}\right)}{x^{2}+y^{2}},} & {(x, y) \neq(0,0)} \\ {0,} & {(x, y)=(0,0)} \\ {\frac{\partial f}{\partial x} \text { and } \frac{\partial f}{\partial y} \text { at }(0,0)}\end{array}\right. $$

Find the value of \(\partial x / \partial z\) at the point \((1,-1,-3)\) if the equation $$x z+y \ln x-x^{2}+4=0$$ defines \(x\) as a function of the two independent variables \(y\) and \(z\) and the partial derivative exists.

You will explore functions to identify their local extrema. Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level curves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. d. Calculate the function's second partial derivatives and find the discriminant \(f_{x x} f_{y y}-f_{x y}^{2}\) . e. Using the max-min tests, classify the critical points found in part (c). Are your findings consistent with your discussion in part (c)? $$\begin{array}{l}{f(x, y)=2 x^{4}+y^{4}-2 x^{2}-2 y^{2}+3,-3 / 2 \leq x \leq 3 / 2} \\ {-3 / 2 \leq y \leq 3 / 2}\end{array}$$

You will explore functions to identify their local extrema. Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level curves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. d. Calculate the function's second partial derivatives and find the discriminant \(f_{x x} f_{y y}-f_{x y}^{2}\) . e. Using the max-min tests, classify the critical points found in part (c). Are your findings consistent with your discussion in part (c)? $$\begin{array}{l}{f(x, y)=5 x^{6}+18 x^{5}-30 x^{4}+30 x y^{2}-120 x^{3}} \\\ {-4 \leq x \leq 3, \quad-2 \leq y \leq 2}\end{array}$$

Let $$f(x, y)=\left\\{\begin{array}{ll}{y^{3},} & {y \geq 0} \\ {-y^{2},} & {y<0}\end{array}\right.$$ Find \(f_{x}, f_{y}, f_{x y},\) and \(f_{y x},\) and state the domain for each partial derivative.
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