Final answer:
The feasible region of the given linear program with constraints 2a+3b≤12 and a−b≥4 is the overlapping area on the coordinate plane where the half-plane below the line 2a+3b≤12 intersects with the half-plane above the line a−b≥4.
Step-by-step explanation:
To find the feasible region for the linear program given the constraints 2a+3b≤12 and a−b≥4.
For the first inequality, 2a+3b≤12, this can be rewritten in slope-intercept form as b = −(2/3)a + 4, which means the corresponding line has a slope of -2/3 and a y-intercept of 4.
For the second inequality, a - b ≥ 4, we can rewrite this as b = a - 4, so the line has a slope of 1 and a y-intercept of -4.
Each inequality represents a half-plane on the coordinate system, and the feasible region will be the area where these half-planes intersect.
Then, since the first inequality is a ≤ relation, shade below the line for 2a+3b≤12.
For the second inequality, since it is a ≥ relation, shade above the line for a - b ≥ 4.
The overlap of these shaded regions is the feasible region.
Question: Consider the following linear program 5a + 2b?
If the constraints of the linear program are 2a+3b≤12 and a−b≥4, find the feasible region on the coordinate plane.