Problem 60 - Calculus Early Transcendentals

Question

For each region \(R\), find the horizontal line \(y=k\) that divides \(R\) into two subregions of equal area. \(R\) is the region bounded by \(y=1-x,\) the \(x\) -axis, and the \(y\) -axis.

Step-by-Step Solution

Answer

Short Answer: To find the horizontal line that bisects a region bounded by y = 1 - x, x-axis, and y-axis, we first calculated the area of the region, which is 1/2. We then determined that the area of each subregion should be 1/4. By using the equation k = 1 - x_A and integrating it with respect to x in the range of 0 to x_B, we found that the line y = k splits the region into two equal parts when k * x_B = 1/4.

Step 1: Sketch the region

To understand the problem better, start by drawing the region \(R\) that is bounded by the x-axis, y-axis, and the line \(y=1-x\). The region is a triangle formed by the intersections of the lines and axes.

Step 2: Calculate the area of the region\(R\)

Since the region is a triangle, its area can be found using the formula \(area= \frac{1}{2}bh\), where \(b\) is the base and \(h\) is the height of the triangle. The base of the triangle is the distance along the x-axis from the y-axis to the point where \(y=1-x\) intersects the x-axis, i.e., \(x=1\). The height of the triangle is the distance from the x-axis to the point where \(y=1-x\) intersects the y-axis, i.e., \(y=1\). Therefore, $$A=\frac{1}{2}(1)(1) = \frac{1}{2}$$.

Step 3: Divide the area in half to find the horizontal line that bisects the region

In order to find the horizontal line that bisects the region, we need to divide the area in half. Since the area of the region is \(\frac{1}{2}\), the area of each subregion should be \(\frac{1}{4}\).

Step 4: Find the equation of the line \(y=k\) that bisects the region into two equal areas

Let's call the points where the line \(y=k\) intersects \(y=1-x\) and the x-axis \(A (x_A,y_A)\) and \(B (x_B, y_B)\), respectively. Since \(y_A = k\) and \(y_B = 0\), we can write \((x_B,y_B) = (x_B, 0)\). Now, we need to find \(x_B\) such that the area of the triangle formed by the points \((0,0)\), \((x_B,0)\), and \((x_B, k)\) is equal to \(\frac{1}{4}\). This area can be calculated as \(\frac{1}{2}(k)(x_B)\). Thus, $$\frac{1}{2}(k)(x_B) = \frac{1}{4}.$$ Since \(y_A=k\) and \(y_A=1-x_A\), we also have \(k=1-x_A\). Using this reality in the above equation, we now have, $$\frac{1}{2}(1-x_A)(x_B)=\frac{1}{4}$$. As point A lies on the line \(y=1-x\), we can write \(y_A=1-x_A\). But as point A's y-coordinate is k, we can substitute k in the equation and get \(k=1-x_A\). To find the value of \(x_B\), we will substitute this \(k\)'value in the area equation and solve for \(x_B\). Therefore, $$\frac{1}{2}(k)(x_B)= \frac{1}{4}$$. $$\frac{1}{2}(1-x_A)(x_B) = \frac{1}{4}$$. Now, we want to find the line that divides the area of the region equally, so we integrate the function \(y=k\) over \(x\) in the range \(0\) to \(x_B\), which will be equal to a quarter of the area of the region. The equation is: $$\int_0^{x_B} k\,dx = \frac{1}{4}$$. Since \(k=1-x_A\), we have: $$\int_0^{x_B}(1-x_A)\,dx = \frac{1}{4}$$. To find the intersection point \(A(x_A, k)\), we realize that it lies on the line \(y = 1 - x\), so \(k = 1 - x_A\). We can now rewrite the integral equation as: $$\int_0^{x_B}(1-(1-k))\,dx = \frac{1}{4}$$. $$\int_0^{x_B}k\,dx = \frac{1}{4}$$. Now integrating and solving for k, we get $$kx\Big|_0^{x_B}=\frac{1}{4}$$$$\implies k\cdot x_B=\frac{1}{4}$$. So we found that the line y=k divides the region \(R\) into two equal parts.