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Chapter 3: Problem 1

Find the area of the region represented by \(\left\\{\begin{array}{l}x+y \leq 2 \\\ x+y \geq 1 \\ x \geq 0 \\ y \geq 0\end{array}\right.\)

Short Answer

Expert verified
Answer: The area of the region is 1 square unit.

Step by step solution

01

Visualize the Inequalities

We have a system of inequalities: 1. \(x+y \leq 2\) 2. \(x+y \geq 1\) 3. \(x \geq 0\) 4. \(y \geq 0\) Let's transform each inequality into an equation to help us visualize the boundaries of the region: 1. \(x+y = 2\) 2. \(x+y = 1\) We will now create a rough sketch of the region bounded by these equations along with the non-negativity conditions.$\\$
02

Find the Corner Points

Now that we have a visual representation of the region, let's find the corner points of the region. 1. Intersect boundary lines: A. \(x+y=1\) and \(x=0\) B. \(x+y=1\) and \(y=0\) C. \(x+y=2\) and \(x=0\) D. \(x+y=2\) and \(y=0\) 2. Solve these intersections to find the corner points: A. \((0,1)\) B. \((1,0)\) C. \((0,2)\) D. \((2,0)\)
03

Calculate the Area of the Region

To calculate the area of the region, we can use the corner points to form two triangles. The first triangle has vertices (0,0), (1,0), and (0,1), and the second triangle has vertices (0,1), (1,0), and (1,2). 1. Calculate the area of the first triangle: Let's use the Shoelace formula to calculate the area: \[ Area = \frac{1}{2} \left|\begin{matrix}0 & 1 & 0 \\0 & 0 & 1\end{matrix} - \begin{matrix}0 & 0 & 1 \\1 & 0 & 0\end{matrix}\right| \\ \] \[ Area = \frac{1}{2}(|0-1|) = \frac{1}{2}(1)=\frac{1}{2} \] 2. Calculate the area of the second triangle: Let's use the Shoelace formula again: \[ Area = \frac{1}{2} \left|\begin{matrix}0 & 1 & 1 \\1 & 0 & 2\end{matrix} - \begin{matrix}1 & 1 & 0 \\0 & 2 &1\end{matrix}\right| \\ \] \[ Area = \frac{1}{2}(|-1+2|) = \frac{1}{2}(1)=\frac{1}{2} \] Now, sum the areas of both triangles to get the total area of the region: \[Area =\frac{1}{2} + \frac{1}{2} = 1\] Thus, the area of the region represented by the given inequalities is 1 square unit.

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Key Concepts

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

Inequalities Visualization
Understanding inequalities is a pivotal skill in mathematics, especially when learning about regions in a two-dimensional plane.

To visualize inequalities like those in the exercise given (i.e., ), it's useful to first transform the inequalities to equations by setting each to an equality. These equations represent boundary lines on the Cartesian plane. Creating a sketch of these lines will help you see the shape of the region you are concerned with. This shape is bounded by:
  • The line where is on or below the line
  • The line where is on or above the line
  • The vertical axis (), representing
  • The horizontal axis (), representing
With these visual markers, you can better perceive the specific area where the inequalities overlap, fundamentally creating a window into the region in question — it is within this window where you'll find the feasible solutions, or in our case, the area that needs to be calculated. After all, a picture is worth a thousand numbers when it comes to grasping mathematical concepts.
Corner Points Method
In calculating the area of a region, identifying the corner points, also known as vertices, is critical. In the step-by-step solution, the corner points were found by analyzing where the boundary lines of the inequalities intersect.

Corner points method is particularly useful in problems dealing with linear inequalities forming a polygonal region. The process goes as follows: you identify each boundary equation and then solve systems of equations that represent their intersections, yielding points like , , , and in our given exercise. Once you have these points, you can effectively outline the shape of your region on the Cartesian plane. Each of these corner points will act as vital references when you proceed to utilize geometric formulas or techniques to calculate the area of the region, building a bridge between algebraic and geometric visualization.
Shoelace Formula
Once corner points of a polygon are known, computing its area can be performed using the Shoelace formula, also known as Gauss's area formula. This clever mathematical technique is named for the cross-multiplication pattern that resembles the lacing of shoes when applied.

The Shoelace formula is used for a simple polygon by taking the coordinates of the vertices. To apply it, you list down the x and y coordinates of all the vertices in a column, then repeat the first vertex at the end of this sequence to close the shape. The area is then calculated by cross-multiplying adjacent y and x terms, summing one diagonal and subtracting the sum of the other diagonal, and finally dividing the absolute value by 2.

The step-by-step solution for our problem uses the Shoelace formula to calculate the areas of two triangles created within the region's boundary. For instance, to find the area of a triangle with vertices , , and , we would organize the coordinates into two columns and apply the formula, leading us to: and thus, the area of units squared. The process is repeated for the second triangle. The sum of these individual areas gives the total area of the region, offering a practical, algebraic approach to solving geometric problems.

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Most popular questions from this chapter

Find area between curves \(\mathrm{y}=\mathrm{x}^{2}\) and \(\mathrm{y}=3 \mathrm{x}-2\) from \(\mathrm{x}=0\) to \(\mathrm{x}=2\).

Show that the area under \(\mathrm{y}=\frac{1}{\mathrm{x}}\) on the interval \([1, a\) a] equals the area under the same curve on \([\mathrm{k}, \mathrm{ka}]\) for any number \(\mathrm{k}>0\).

Which of the following statements are true? (i) If \(\mathrm{C}>0\) is a constant, the region under the line \(\mathrm{y}=\mathrm{C}\) on the interval \([\mathrm{a}, \mathrm{b}]\) has area \(\mathrm{A}=\mathrm{C}(\mathrm{b}-\mathrm{a})\) (ii) IfC> 0 is a constant and \(\mathrm{b}>\mathrm{a} \geq 0\), the region under the line \(\mathrm{y}=\mathrm{Cx}\) on the interval \([\mathrm{a}, \mathrm{b}]\) has area \(\mathrm{A}=\frac{1}{2} \mathrm{C}(\mathrm{b}-\mathrm{a})\) (iii) Theregion under the parabolay \(=\mathrm{x}^{2}\) on the interval \([\mathrm{a}, \mathrm{b}]\) has area less than \(\frac{1}{2}\left(\mathrm{~b}^{2}+\mathrm{a}^{2}\right)(\mathrm{b}-\mathrm{a})\). (iv) The region under the curve \(\mathrm{y}=\sqrt{1-\mathrm{x}^{2}}\) on the interval \([-1,1]\) has area \(\mathrm{A}=\frac{\pi}{2}\). (v) Let f be a function that satisfies \(\mathrm{f}(\mathrm{x}) \geq 0\) for \(\mathrm{x}\) in the interval \([a, b]\). Then the area under the curve \(\mathrm{y}=\mathrm{f}^{2}(\mathrm{x})\) on the interval \([\mathrm{a}, \mathrm{b}]\) mustalways be greater than the area under \(\mathrm{y}=\mathrm{f}(\mathrm{x})\) on the same interval. (vi) If f is even and \(\mathrm{f}(\mathrm{x}) \geq 0\) throughout the interval \([-a, a]\), then the area under the curve \(y=f(x)\) on this interval is twice the area under \(\mathrm{y}=\mathrm{f}(\mathrm{x})\) on \([0, a]\).

Compute the area of the curvilinear trapezoid bounded by the \(x\)-axis and the curve \(y=x-x^{2} \sqrt{x}\).

Find the area of the closed figure bounded by the curves \(\mathrm{y}=2-|2-\mathrm{x}|\) and \(\mathrm{y}=\frac{3}{|\mathrm{x}|}\)
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