Calculat the area of the region enclosed by the graph: abs value(4x-12)+ abs value(10y-20)=60?

Answer 1

The boundary descibed gives rise to 4 cases:
1.) (4x - 12) and (10y - 20) are both positive
2.) (4x - 12) and (10y - 20) are both negative
3.) (4x - 12) is positive and (10y - 20) is negative
4.) (4x - 12) is negative and (10y - 20) is positive
Case 1.) 4x - 12 > 0 & 10y - 20 > 0
===> 4x > 12 & 10y > 20
===> x > 3 & y > 2
This set of inequalities denotes a quadrant of the x-y plane in which the following equation prevails:
4x - 12 + 10y - 20 = 60
4x + 10 y = 92
10y = -4x + 92
y = (-2/5)x + 46/5
This line intersects the borderline of the quadrant at:
x = 3 ===> y = (-2/5)(3) + 46/5 = 8 the point (3, 8)
y = 2 ===> 2 = (-2/5)x + 46/5; x = 18 the point (18, 2)
Case 2.) 4x - 12 < 0 & 10y - 20 < 0
===> x < 3 & y < 2
In this quadrant the following equation prevails:
-(4x- 12) - (10y - 20) = 60
-4x + 12 - 10y + 20 = 60
-10y = 4x + 28
y = (-4/10)x - 28/10
y = (-2/5)x - 14/5
This line intersects the borderline of the quadrant at:
x = 3 ===> y = (-2/5)(3) - 14/5 = -20/5 = -4 the point (3, -4)
y = 2===> 2 = (-2/5)x - 14/5 (2/5)x = -10/5 - 14/5 point (-12, 2)
Working the other two cases yields the same intersection points so the solution set of the original equation is the rhombus formed by connecting the points:
(3, 8), (18, 2), (3, -4), and (-12,2). An inspection of the graph shows that the rhombus consists of 4 congruent triangles with base 15 and height 6.
A = 4(1/2)bh
A = 4(1/2)(15)(6) = 180square units