Question

Shade the feasible region for the inequalities
\( 6x + 4y \leq 120 \), \( 3x + 10y \leq 180 \), \( x, y \geq 0 \) in a rough figure.

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

When solving linear inequalities graphically, first graph the boundary lines, then shade the region where all the inequalities hold. Always check the corner points to determine if they satisfy all inequalities.

Answer 1

The Correct Option is B

We are given the following inequalities: \[ 6x + 4y \leq 120, \quad 3x + 10y \leq 180, \quad x, y \geq 0 \] Step 1: Rewriting the inequalities - The first inequality \( 6x + 4y \leq 120 \) can be rearranged as: \[ y \leq \frac{120 - 6x}{4} = 30 - 1.5x \] This represents a line with slope \(-1.5\) and y-intercept \(30\). - The second inequality \( 3x + 10y \leq 180 \) can be rearranged as: \[ y \leq \frac{180 - 3x}{10} = 18 - 0.3x \] This represents a line with slope \(-0.3\) and y-intercept \(18\). Step 2: Graphing the inequalities
- The line \( y = 30 - 1.5x \) intersects the y-axis at \( (0, 30) \) and the x-axis at \( (20, 0) \).
- The line \( y = 18 - 0.3x \) intersects the y-axis at \( (0, 18) \) and the x-axis at \( (60, 0) \).
- The feasible region is bounded by these lines and the axes. Step 3: Analyzing the options - Option (B) shows the correct feasible region where both inequalities are satisfied along with the conditions \( x, y \geq 0 \). Thus, the correct answer is option B.