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Chapter 3: Problem 53

Graph the indicated functions. An airline requires that any carry-on bag has total dimensions (length \(+\) width \(+\) height) that do not exceed 45 in. For a carryon that just meets this requirement and \(h\) as a length that is twice the width, express the volume \(V\) as a function of the width \(w\). Draw the graph of \(V=f(w)\)

Short Answer

Expert verified
The volume function is \(V(w) = 90w^2 - 6w^3\). Graph this function for \(0 < w \leq 15\).

Step by step solution

01

Understanding the Problem

We have a carry-on bag with a total dimension (length + width + height) of 45 inches. The length is twice the width, and we need to express the volume as a function of the width.
02

Setting Up the Equations

Let the width be denoted by \(w\), the length by \(l\), and the height by \(h\). According to the problem, \(l = 2w\). Therefore, the sum of the dimensions is given by \(l + w + h = 45\), which can be rewritten as \(2w + w + h = 45\). Thus, \(3w + h = 45\).
03

Solving for the Height

From \(3w + h = 45\), we isolate \(h\): \(h = 45 - 3w\). We now have expressions for all dimensions in terms of \(w\).
04

Finding the Volume Function

The volume \(V\) of a rectangular prism is given by the product of its dimensions: \(V = lwh\). Substituting the expressions for \(l\) and \(h\) in terms of \(w\), we get: \(V = (2w)(w)(45 - 3w)\). Simplifying this, \(V = 2w^2(45 - 3w)\).
05

Simplifying the Volume Function

Expand and simplify the volume function: \(V = 2w^2 \times 45 - 2w^2 \times 3w = 90w^2 - 6w^3\). Therefore, the volume as a function of the width is \(V(w) = 90w^2 - 6w^3\).
06

Graphing the Function

To graph \(V(w) = 90w^2 - 6w^3\), plot \(w\) on the x-axis and \(V(w)\) on the y-axis. Identify critical points, such as when the derivative is zero, to find maxima or minima. The domain of \(w\) is restricted such that \(3w \leq 45\), hence \(w \leq 15\). Evaluate points of interest within this domain.

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Key Concepts

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

graphing functions
Graphing functions is an essential part of understanding mathematical relationships between variables. In this exercise, we focus on graphing the volume function, which represents how the volume of a carry-on bag changes with its width.
To graph the function \( V(w) = 90w^2 - 6w^3 \), you need to understand how changes in the value of \( w \), the width, impact the volume. Here, the goal is to plot width \( w \) on the x-axis and the volume \( V(w) \) on the y-axis.
The first step is to determine the domain. Since the total dimension cannot exceed 45, for this problem, \( 3w \leq 45 \) implies \( w \leq 15 \). Hence, you'll only consider \( w \) values from 0 to 15.
  • Observe the function's degree and leading coefficient. With \(-6w^3\) as the term with the highest degree, expect a downward-opening cubic curve.
  • Calculate critical points using the derivative to find local maxima or minima.
  • Evaluate specific width values to sketch the curve, considering \( w = 0, 5, 10, \) and \( 15 \).
expressions in terms of variables
Expressions in terms of variables are fundamentally about using symbols to represent numbers, making them flexible and adaptable to various situations.
In this particular problem, you express all dimensions of the carry-on bag using just one variable, \( w \). The length \( l \) is twice the width, given by \( l = 2w \), while the height \( h \) is derived from the total dimension constraint: \( h = 45 - 3w \).

Using expressions simplifies understanding and calculation:
  • It reduces complexity, making it easier to perform calculations by substituting different values for \( w \).
  • You can generalize the problem to other similar scenarios by changing the variable expressions accordingly, without needing to start from scratch.
Understanding how to express dimensions with variables is key in creating mathematical models that mimic real-world constraints.
problem-solving steps
Solving problems systematically through a series of well-defined steps helps break down complex challenges into manageable parts. Here's a quick overview following the provided solution:
  • Step 1: Understand the Problem - Comprehend all given constraints and what needs to be expressed or calculated, such as expressing the volume in terms of width.
  • Step 2: Set Up the Equations - Translate the constraints into mathematical expressions and equations using known relationships, like setting \( l = 2w \).
  • Step 3: Solve for Variables - Derive expressions for remaining variables, such as solving for height: \( h = 45 - 3w \).
  • Step 4: Develop the Function - Formulate the volume function \( V(w) \) using the expressions from earlier steps.
  • Step 5: Graph the Function - Visualize the output by plotting the function and identifying key features such as maximums or minimums.
This structured approach is invaluable, especially when dealing with complex calculus problems or functions.
calculus applications
Calculus is a powerful tool used extensively for analyzing functions and their behavior. In this exercise, calculus can help identify important features of the volume function \( V(w) = 90w^2 - 6w^3 \).
By applying calculus concepts, such as derivation, you can:
  • Find critical points by setting the derivative to zero, \( V'(w) = 180w - 18w^2 = 0 \), to solve for \( w \).
  • Understand the behavior of the function around these points to determine if they are maxima or minima.
  • Determine intervals of increase or decrease by analyzing the sign of \( V'(w) \).
These techniques provide insights into how the volume changes with respect to the width and help optimize the space within constraints. Understanding such calculus applications enables you to solve more advanced problems efficiently.

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Most popular questions from this chapter

Solve the given problems. The length \(L\) (in \(\mathrm{ft}\) ) of a cable hanging between equal supports \(\left.100 \mathrm{ft} \text { apart is } L=100\left(1+0.0003 s^{2}\right), \text { where } s \text { is the sag (in } \mathrm{ft}\right)\) in the middle of the cable. Because \(L=f(s),\) find \(f(15)\).

Graph the indicated functions. The resistance \(R\) (in \(\Omega\) ) of a resistor as a function of the temperature \(T\) (in \(^{\circ} \mathrm{C}\) ) is given by \(R=250(1+0.0032 T) .\) Plot \(R\) as a function of \(T\)

Graph the indicated functions. A guideline of the maximum affordable monthly mortgage \(M\) on a home is \(M=0.25(I-E),\) where \(I\) is the homeowner's monthly income and \(E\) is the homeowner's monthly expenses. If \(E=\$ 600,\) graph \(M\) as a function of \(I\) for \(I=\$ 2000\) to \(I=\$ 10,000\)

Graph the given functions. $$y=x^{3}-x^{2}$$

Represent the data graphically. The amount of material necessary to make a cylindrical gallon container depends on the diameter, as shown in this table: $$\begin{array}{l|c|c|c|c|c|c|c} \text {Diameter (in.)} & 3.0 & 4.0 & 5.0 & 6.0 & 7.0 & 8.0 & 9.0 \\ \hline \text {Material }\left(\text { in. }^{2}\right) & 322 & 256 & 224 & 211 & 209 & 216 & 230 \end{array}$$
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