Consider the region in Quadrant 1 totally bounded by the 4 lines: x = 3, x = 9, y = 0, and y = mx (where m is positive). Determine the value of c such that the vertical line x = c bisects the area of that totally bounded region.
Needless to say, your first task should be to draw a representative diagram for the problem that has been described. Also, after determining the value of c, be sure to comment on the (probably) surprising non-contributing factor in this problem.
Bonus: What if the problem were modified such that a horizontal line, y = k was to be the area bisector of the totally bounded region. Determine what the value of k would be in that case.
If we have a trapezoid with parallel sides of a & b then the length of the parallel line that bisects the trapezoid is = rms(a,b)= [(a^2+b^2)/2]^(1/2)
In this case the parallel lines are of lengths 3*m & 9*m and thus the length of the bisector// to the y-axis=m*(45)^(1/2)
Thus, mx=m*(45)^(1/2) or the vertical line x = (45)^(1/2) bisects the given trapezoid.
The surprising non-contributory factor is, m = the slope of the line y=mx.
In the modified case, area of the trapezoid = (3m+9m)*6/2 =9*(m+3)
Thus, k*6 =9*(m+3)/2 or k=(3/4)*(m+3)
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To solve the problem, let's first draw a representative diagram of the region described in Quadrant 1.
|
9 ____________________
| \ \
| \ \
| \_______________\
| / /
| / /
3 |_____/_______________/ (3,0)
| x = 3
| x = 9
| y = 0
|___ y = mx
In the diagram above, the lines x = 3 and x = 9 form the horizontal boundaries, while the lines y = 0 and y = mx form the vertical boundaries.
To determine the value of c such that the vertical line x = c bisects the area of the region, we need to find the equation of the line y = mx and then calculate the area of the region on both sides of the line x = c.
First, let's find the equation of the line y = mx. We know that the line passes through the point (3,0), so we can use this point to determine the value of m. Using the point-slope form of a linear equation, we have:
y - y1 = m(x - x1)
y - 0 = m(x - 3)
y = mx - 3m
Now that we have the equation y = mx - 3m, we can find the value of m by substituting the coordinates of the other point of intersection (9, mx) into the equation. We get:
mx = m*9 - 3m
mx = 9m - 3m
mx = 6m
x = 6
Therefore, the equation of the line y = mx is y = 6x.
To determine the value of c, we need to calculate the area of the region on both sides of the line x = c. We'll divide the region into two triangles and subtract the areas of the smaller triangles from the larger ones.
The total area of the region can be calculated as the difference between the area under the line y = 6x and the area under the line y = 0 between x = 3 and x = 9.
Area of the region = integral[3 to 9](6x - 0)dx
= integral[3 to 9]6xdx
= 6 * integral[3 to 9]xdx
= 6 * [(1/2)x^2] [3 to 9]
= 6 * [(1/2)(9^2) - (1/2)(3^2)]
= 6 * [(1/2)(81) - (1/2)(9)]
= 6 * [(81/2) - (9/2)]
= 6 * (72/2)
= 6 * 36
= 216
To bisect the area, the areas on both sides of x = c should be equal. So, we need to find the value of c such that the area to the left of x = c is half of the total area, i.e., 216/2 = 108.
Let's calculate the area to the left of x = c:
Area to the left of x = c = integral[3 to c](6x - 0)dx
= integral[3 to c]6xdx
= 6 * integral[3 to c]xdx
= 6 * [(1/2)x^2] [3 to c]
= 6 * [(1/2)(c^2) - (1/2)(3^2)]
= 6 * [(1/2)(c^2) - (1/2)(9)]
To find the value of c, we need to solve the equation:
6 * [(1/2)(c^2) - (1/2)(9)] = 108
Simplifying the equation:
3c^2 - 27 = 108
3c^2 = 135
c^2 = 45
c = sqrt(45)
c = 3sqrt(5)
So, the value of c such that the vertical line x = c bisects the area of the region is c = 3sqrt(5).
Now, let's consider the modified problem where a horizontal line y = k bisects the totally bounded region.
To calculate the value of k, we need to calculate the area of the region above y = k and below y = mx, and then find the value of k such that the area above y = k is half of the total area.
We can calculate this area by subtracting the area under y = k from the area under y = mx.
Area below y = mx = integral[3 to 9](mx - 0)dx
= m * integral[3 to 9]xdx
= m * [(1/2)x^2] [3 to 9]
= m * [(1/2)(9^2) - (1/2)(3^2)]
= m * [(1/2)(81) - (1/2)(9)]
= m * [(81/2) - (9/2)]
= m * (72/2)
= 36m
Area above y = k = integral[3 to 9](k - mx)dx
= k * integral[3 to 9]dx - m * integral[3 to 9]xdx
= k * (9 - 3) - 36m
= 6k - 36m
To bisect the area, the areas above and below y = k should be equal. So, we need to find the value of k such that the area above y = k is half of the total area, i.e., 36m/2 = 18m.
Setting the areas equal to each other:
36m = 6k - 36m
Simplifying the equation:
72m = 6k
k = 12m
Therefore, in the modified problem, the value of k that would bisect the totally bounded region when the line y = k is parallel to the x-axis would be k = 12m.