Math Problem Statement

If the area(in sq, units) of the region {(x, y) : 12 ≤4x,x+y≤1,x20, y20} is av2 +b, then a-b=

Solution

To solve the given problem, we need to find the area of the region described by the inequalities:

1. & \quad 12 \leq 4x \\ 2. & \quad x + y \leq 1 \\ 3. & \quad x \geq 0 \\ 4. & \quad y \geq 0 \end{aligned}$$ ### Step-by-step Solution: #### Step 1: Simplify the inequalities 1. From $$12 \leq 4x$$, we divide through by 4: $$x \geq 3.$$ 2. The inequality $$x + y \leq 1$$ represents a line: $$y \leq 1 - x.$$ 3. The other inequalities ($$x \geq 0$$ and $$y \geq 0$$) ensure the region is in the first quadrant. Thus, the region is bounded by: - $$x \geq 3$$, - $$y \leq 1 - x$$, - $$x \geq 0$$, and - $$y \geq 0$$. #### Step 2: Determine the vertices of the region The vertices are obtained by solving intersections of the boundaries: 1. Intersection of $$x = 3$$ and $$y = 1 - x$$: $$y = 1 - 3 = -2.$$ This point is invalid because $$y \geq 0$$. ####It seems I misunderstood the interpretation of the inequalities initially. Let’s carefully analyze the problem step-by-step again to determine the correct bounded region and area. We are given the region described by these inequalities: 1. $$12 \leq 4x$$ $$\Rightarrow x \geq 3$$. 2. $$x + y \leq 1$$. 3. $$x \geq 0$$. 4. $$y \geq 0$$. We aim to find the area of the bounded region formed by these inequalities. --- ### Step 1: Region Description - $$x \geq 3$$: This indicates that the region is to the right of the vertical line $$x = 3$$. - $$x + y \leq 1$$: This represents the region below the line $$y = 1 - x$$, which intersects the axes at $$(1, 0)$$ and $$(0, 1)$$. - $$x \geq 0$$: This ensures $$x$$ is non-negative (right of the $$y$$-axis). - $$y \geq 0$$: This ensures $$y$$ is non-negative (above the $$x$$-axis). However, $$x \geq 3$$ restricts the feasible region to the right of $$x = 3$$. Notice that $$x = 3$$ lies **outside** the line $$x + y \leq 1$$, so the inequalities define no **finite region** in this case. --- ### Step 2: Recheck Problem Constraints If this is the intended problem setup, the region described has **no finite area**, as the constraints $$x \geq 3$$ and $$x + y \leq 1$$ are incompatible. If the problem or region requires re-checking, we can revisit. Would you like to verify the problem details? Alternatively, feel free to provide clarifications or upload a diagram for further analysis!

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Math Problem Analysis

Mathematical Concepts

Linear Inequalities
Geometric Regions
Area Calculation

Formulas

Area of a triangle = 0.5 × base × height

Theorems

Intersection of linear boundaries forms vertices of a polygon

Suitable Grade Level

Grades 10-12

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