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Chapter 4: Problem 1

You are asked to express one variable as a function of another. Be sure to state a domain for the function that reflects the constraints of the problem. (a) The perimeter of a rectangle is \(16 \mathrm{cm} .\) Express the area of the rectangle in terms of the width \(x\). Suggestion: First reread Example \(1 .\) (b) The area of a rectangle is \(85 \mathrm{cm}^{2} .\) Express the perimeter as a function of the width \(x\).

Short Answer

Expert verified
(a) Area: \( A(x) = x(8-x) \), domain: \( 0 < x < 8 \). (b) Perimeter: \( P(x) = 2x + \frac{170}{x} \), domain: \( x > 0 \).

Step by step solution

01

Define Variables and Equations for Part (a)

For a rectangle with a perimeter of \( P = 16 \) cm, let the width be \( x \) cm and the length be \( y \) cm. The perimeter relation is given by \( 2x + 2y = 16 \). Simplify this equation to find \( y \) in terms of \( x \): \( y = 8 - x \). The area of the rectangle \( A \) is given by \( A = xy \). Substitute \( y = 8 - x \) to express the area in terms of \( x \): \( A(x) = x(8-x) \).
02

Determine Domain for Part (a)

The width \( x \) of the rectangle must be non-negative and also less than the length, so \( 0 < x < 8 \). This ensures that both width and length are positive, maintaining the physical meaning of the rectangle's dimensions.
03

Express Perimeter in Terms of Width for Part (b)

Given the area \( A = 85 \) cm\(^2\), the relationship \( xy = 85 \) holds, where \( x \) is the width and \( y \) is the length. Solve for \( y \) in terms of \( x \): \( y = \frac{85}{x} \). Substitute into the perimeter formula \( P = 2x + 2y \): \( P(x) = 2x + 2\left(\frac{85}{x}\right) = 2x + \frac{170}{x} \).
04

Determine Domain for Part (b)

The width \( x \) must be positive (\( x > 0 \)) since it represents a physical dimension. Also, \( x \) should be such that the length \( y = \frac{85}{x} \) is positive, ensuring both dimensions remain physically meaningful for the rectangle.

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Key Concepts

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

Perimeter of Rectangle
The perimeter of a rectangle is the total distance around its outer edges. It can be calculated by adding the lengths of all four sides. In mathematical terms, the formula is:
  • Perimeter, \( P = 2x + 2y \)
Here, \( x \) is the width, and \( y \) is the length of the rectangle.
When you know the perimeter, you can derive relationships between width and length. For example, if the perimeter is 16 cm, you can express the length in terms of the width as follows:
  • \( 2x + 2y = 16 \)
  • \( y = 8 - x \)
This equation comes from simplifying and rearranging the perimeter formula. Understanding these relationships helps when solving problems with limited information on rectangle dimensions.
Area of Rectangle
The area of a rectangle represents the amount of space it occupies on a flat surface. It is determined by multiplying the width by the length. The formula is:
  • Area, \( A = xy \)
In the given exercise, we express the area as a function of the width after recording the relationship between width and length:
  • If \( y = 8 - x \), then \( A(x) = x(8-x) \)
This function, \( A(x) = x(8-x) \), allows us to calculate the area for any permissible width of the rectangle.
By substituting values within the domain, you can find the area for any given width while keeping in mind that the rectangle's dimensions must remain valid.
Domain Constraints
In mathematics, domain constraints ensure that we only consider values that make sense for our function. For rectangles, physical dimensions like width and length must be positive and logical.
When expressing area and perimeter as functions of width, it鈥檚 crucial to identify these limits.### For Part (a):The width \( x \) must satisfy:
  • Greater than 0, because negative width isn't realistic: \( x > 0 \)
  • Less than 8, ensuring the length \( y = 8 - x \) remains positive: \( x < 8 \)
### For Part (b):The width \( x \) should be:
  • Greater than 0 for the same reason as above: \( x > 0 \)
Since the area is fixed at 85 cm\(^2\), the domain ensures that any width value used makes physical and mathematical sense, keeping both dimensions non-negative and meaningful. Without careful attention to these constraints, solutions could become invalid or nonsensical.

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Most popular questions from this chapter

The following table and scatter plot show global coal consumption for the years \(1990-1995\). $$\begin{array}{cc} \hline \text { Year } x & \text { Coal consumption } y \\ \hline x=0 \leftrightarrow 1990 & \text { (billion tons) } \\ \hline 0 & 3.368 \\ 1 & 3.285 \\ 2 & 3.258 \\ 3 & 3.243 \\ 4 & 3.261 \\ 5 & 3.311 \\ \hline \end{array}$$ (GRAPH CAN'T COPY) (a) Use a graphing utility to find a quadratic model for the data. Then use the model to make estimates for global coal consumption in 1989 and 1996 (b) Use the following information to show that, in terms of percentage error, the 1996 estimate is better than the 1989 estimate, but in both cases the percentage error is less than \(2 \% .\) The actual figures for coal consumption in 1989 and 1996 are 3.408 and 3.428 billion tons, respectively. (c) Use the model to project worldwide coal consumption in \(1998 .\) Then show that the percentage error is more than \(9 \%,\) given that the actual 1998 consumption was 3.329 billion tons.

(a) Determine the \(x\) - and \(y\) -intercepts and the excluded regions for the graph of the given function. Specify your results using a sketch similar to Figure \(16(a) .\) In Exercises \(31-34\) you will first need to factor the polynomial. (b) Graph each function. $$y=(x-2)(x-1)(x+1)$$

Suppose we have a table of \(x\) -y values with the \(x\) -values equally spaced. This exercise shows that if the first differences of the \(y\) -values are constant, then the table can be generated by a linear function. Note: To show that the data points are generated by a linear function, it is enough to show that the points all lie on one nonvertical line. Consider the following table with the \(x\) -values equally spaced by an amount \(h \neq 0 .\) (We are assuming that \(h\) is nonzero to guarantee that the three \(x\) -values are distinct.) \begin{tabular}{llll} \hline\(x\) & \(a\) & \(a+h\) & \(a+2 h\) \\ \(y\) & \(y_{1}\) & \(y_{2}\) & \(y_{3}\) \\ \hline \end{tabular} lues are con- is a constant. he two data joining the date data three data Assuming that the first differences of the \(y\) -values a stant, we have \(y_{2}-y_{1}=y_{3}-y_{2}=k\), where \(k\) is a con (a) Check that the slope of the line joining the two dat points \(\left(a, y_{1}\right)\) and \(\left(a+h, y_{2}\right)\) is \(k / h\). (b) Likewise, check that the slope of the line joining the two data points \(\left(a+h, y_{2}\right)\) and \(\left(a+2 h, y_{3}\right)\) is \(k / h\). From parts (a) and (b), we conclude that the three data points lie on a nonvertical line, as required. (The slope \(k / h\) is well defined because we assumed \(h \neq 0 .\) )

Graph the functions. Note: In each case, the graph crosses its horizontal asymptote once. To find the point where the rational function \(y=f(x)\) crosses its horizontal asymptote \(y=k,\) you \(1 /\) need to solve the equation \(f(x)=k\). $$y=\frac{(x-1)(x-3)}{(x+1)^{2}}$$

Graph the functions. Note: In each case, the graph crosses its horizontal asymptote once. To find the point where the rational function \(y=f(x)\) crosses its horizontal asymptote \(y=k,\) you \(1 /\) need to solve the equation \(f(x)=k\). $$y=\frac{(x+1)^{2}}{(x-1)(x-3)}$$
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