A region of the plane is defined by:
|y-2x| + |y+2x| + |y-3| ≤ 9
What is the area of this region?
The absolute value function changes from being the positive of its argument to the negative of its argument when that argument changes from positive to negative, so for example, when y-2x is positive, then abs(y-2x)=y-2x, but when y-2x is negative, abs(y-2x) = 2x-y.
Other than the absolute value functions, the relationships of x and y are linear. It's just that they are different linear relations in different regions of the Cartesian plane. These regions (not the region of the figure whose area is to be found) are bounded by the lines where each of the arguments to the absolute value function are zero: y=2x, y=-2x, and y=3. These lines divide the plane into seven regions--one central triangle ("inverted") and six surrounding regions.
In the central triangle, y<3, y>2x and y>-2x. The inequality there is y-2x + y+2x + 3-y <= 9, or y + 3 <= 9, or y <= 6, which of course it is in this region as y <= 3. So this entire central region is within the area to be found.
In the region above this, where y>3, y>2x and y>-2x, the inequality is 3y-3 <=9 or 3y <= 12 or y <= 4. So y=4 is an upper boundary of the area to be found, extending from (-2,4) on the left to (2,4) on the right, where y=4 meets y=-2x and y=2x respectively.
In the region to the right of this, where y>3, y<2x and of course y>-2x, the inequality is 4x+y-3 <= 9, or 4x+y <= 12. So the line y=12-4x forms an upper-right boundary of the desired area, extending from (2,4) to (2.25,3).
In the region below this, where y<3, y<2x and y>-2x, the inequality is 4x-y+3 <= 9 or 4x-y <= 12, so a lower-right boundary of the area to be found is y=4x-12. This extends from (2.25,3) to (1,-2), where this line meets the boundary of the region.
When y<3, y<2x and y<-2x, the inequality is -3y+3 <= 9 or 3y >= -6 or y >= -2. So y=-2 forms the bottom boundary of our area between (1,-2) on the right and (-1,-2) on the left.
Symmetrically we see the remaining vertex is at (-2.25,3).
This irregular (but symmetric left-right) hexagon can be broken into two rectangles, two small right triangles and two larger right triangles.
The rectangle from (-2,4) to (2,3) has area 4.
The rectangle from (-1,3) to (1,-2) has area 10.
The smaller triangles, (2,4) to (2.25,3) and (-2,4) to (-2.25,3) add up to 1/4 in area.
The larger triangles, (2.25,3) to (1,-2) and (-2.25,3) to (-1,-2) add up to 6.25 in area.
The area adds up to 20.5.
Posted by Charlie
on 2004-12-01 20:17:26