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Chapter 4: Problem 57

Give an example of: The sides of a rectangle with perimeter \(20 \mathrm{cm}\) and area smaller than \(10 \mathrm{cm}^{2}.\)

Short Answer

Expert verified
Sides: 9 cm and 1 cm; Area: 9 cm虏 (smaller than 10 cm虏); Perimeter: 20 cm.

Step by step solution

01

Recall the formulas

To solve this problem, we first need to recall the formulas for the perimeter and area of a rectangle. The perimeter of a rectangle with sides of lengths \( l \) and \( w \) is given by the formula: \( P = 2l + 2w \). The area is given by: \( A = lw \).
02

Set up the perimeter equation

We know that the perimeter is 20 cm, so we substitute this into the perimeter formula: \( 2l + 2w = 20 \). Simplifying the equation, we get: \( l + w = 10 \).
03

Express one side in terms of the other

From the equation \( l + w = 10 \), we can express \( l \) in terms of \( w \): \( l = 10 - w \).
04

Set up the area condition

We need the area \( A = lw \) to be smaller than 10 cm虏. Substituting \( l = 10 - w \) into this area equation gives: \( A = (10-w)w < 10 \).
05

Solve the inequality

The inequality \( (10-w)w < 10 \) simplifies to \( 10w - w^2 < 10 \) or \( -w^2 + 10w - 10 < 0 \). By solving the quadratic equation \( -w^2 + 10w - 10 = 0 \), we find the solutions \( w = 1.138 \) and \( w = 8.862 \). The values for \( w \) must be between these two roots.
06

Choose specific side lengths

Choose values for \( w \) between 1.138 and 8.862. For example, let \( w = 2 \). Then \( l = 10 - 2 = 8 \). Check the area: \( A = 8 \times 2 = 16 \), which is too large. Try again with \( w = 1.5 \), giving \( l = 8.5 \), producing: \( A = 1.5 \times 8.5 = 12.75 \), still too large. Finally, consider \( w = 3 \), and \( l = 7 \), resulting in \( A = 3 \times 7 = 21 \) 鈥� another unsuitable value. In the process of elimination, the only combination that works is \( w = 1 \), giving \( l = 9 \). This results in \( A = 1 \times 9 = 9 \).
07

Verify all conditions

The perimeter of the rectangle with sides 9 cm and 1 cm is \( 2(9) + 2(1) = 18 + 2 = 20 \) cm, which meets the perimeter requirement. The area \( 9 \times 1 = 9 \) cm虏 is smaller than 10 cm虏 and also meets the condition.

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Key Concepts

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

Understanding Geometry and Rectangles
In geometry, a rectangle is a four-sided shape, often featured prominently due to its practical applications and simplicity. The key characteristics of a rectangle include its length and width, with opposite sides being equal and all angles at 90 degrees.
Understanding the properties of a rectangle will help to solve various mathematical problems, such as calculating the perimeter and area.
For a rectangle:
  • Perimeter: The perimeter is the total length around the rectangle. It is calculated using the formula: \( P = 2l + 2w \), where \( l \) is the length, and \( w \) is the width.
  • Area: The area measures the surface covered by the rectangle and is given by the formula: \( A = lw \).
These formulas are foundational in geometry, providing a means to solve a multitude of problems and applications involving rectangles.
Solving Quadratic Inequalities
Quadratic inequalities involve expressions in the form of \( ax^2 + bx + c < 0 \) or \( ax^2 + bx + c > 0 \). These mathematical expressions describe a range of values rather than a specific number, indicating where a parabola sits relative to the x-axis.
To solve a quadratic inequality:
  • First, solve the related quadratic equation \( ax^2 + bx + c = 0 \) to find critical points 鈥� these are often called roots or zero points.
  • Use these roots to define intervals and determine where the inequality holds true.
  • Check test points within each interval to see whether they satisfy the inequality.
In our rectangle problem, we solve the quadratic inequality \( -w^2 + 10w - 10 < 0 \) to determine the valid range of values for the width \( w \) where the area stays under a certain threshold, ensuring all criteria are met regarding its size and shape.
Using Formula Manipulation
Formula manipulation is a key skill in mathematics, allowing you to rearrange and transform equations to suit your needs. This skill is invaluable when finding unknown values in word problems, like determining dimensions of a shape given other constraints.
When tackling problems of this nature:
  • Start by isolating the variable you want to solve for; for example, derived from the perimeter equation \( l + w = 10 \), you can express \( l \) in terms of \( w \), thereby reducing unknown variables.
  • Substitute these expressions into other formulas, such as the area formula, to further simplify or solve the problem.
  • Adjust the formula as needed to test potential solutions against any additional conditions or limits, as shown with the rectangle's width and length.
This systematic approach helps in breaking down complex problems into more manageable parts, finding accurate solutions, like achieving a rectangle's area less than 10 cm虏 while maintaining the correct perimeter.

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Most popular questions from this chapter

In a romantic relationship between Angela and Brian, who are unsuited for each other, \(a(t)\) represents the affection Angela has for Brian at time \(t\) days after they meet, while \(b(t)\) represents the affection Brian has for Angela at time \(t .\) If \(a(t)>0\) then Angela likes Brian; if \(a(t)<0\) then Angela dislikes Brian; if \(a(t)=0\) then Angela neither likes nor dislikes Brian. Their affection for each other is given by the relation \((a(t))^{2}+(b(t))^{2}=c\) where \(c\) is a constant. (a) Show that \(a(t) \cdot a^{\prime}(t)=-b(t) \cdot b^{\prime}(t)\) (b) At any time during their relationship, the rate per day at which Brian's affection for Angela changes is \(b^{\prime}(t)=-a(t) .\) Explain what this means if Angela (i) Likes Brian, (ii) Dislikes Brian. (c) Use parts (a) and (b) to show that \(a^{\prime}(t)=b(t) .\) Explain what this means if Brian (i) Likes Angela, (ii) Dislikes Angela. (d) If \(a(0)=1\) and \(b(0)=1\) who first dislikes the other?

For positive \(a, b,\) the potential energy, \(U,\) of a particle is $$U=b\left(\frac{a^{2}}{x^{2}}-\frac{a}{x}\right) \quad \text { for } x>0$$ (a) Find the intercepts and asymptotes. (b) Compute the local maxima and minima. (c) Sketch the graph.

A water tank is in the shape of an inverted cone with depth 10 meters and top radius 8 meters. Water is flowing into the tank at 0.1 cubic meters/min but leaking out at a rate of \(0.004 h^{2}\) cubic meters/min, where \(h\) is the depth of the water in meters. Can the tank ever overflow?

Given that \(f^{\prime}(x)\) is continuous everywhere and changes from negative to positive at \(x=a,\) which of the following statements must be true? (a) \(a\) is a critical point of \(f(x)\) (b) \(f(a)\) is a local maximum (c) \(f(a)\) is a local minimum (d) \(f^{\prime}(a)\) is a local maximum (e) \(f^{\prime}(a)\) is a local minimum

Let \(g(x)=x-k e^{x},\) where \(k\) is any constant. For what value(s) of \(k\) does the function \(g\) have a critical point?
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