1. Show graphically the region \( R \) which satisfies the set of inequalities: \( 2 x+3 y \leq 26 \) \( x+2 y \) \( \begin{aligned} & \leq 16 \\ x & \geq 0 \\ y & \geq 0\end{aligned} \)

It seems that the system does not currently support plotting. However, I can guide you through the process of graphing the region \( R \) that satisfies the given inequalities step by step. ### Step 1: Graph the Boundary Lines 1. **For the inequality \( 2x + 3y \leq 26 \)**: - First, find the boundary line by setting the equation to equality: \[ 2x + 3y = 26 \] - To find the intercepts: - Set \( x = 0 \): \[ 3y = 26 \implies y = \frac{26}{3} \approx 8.67 \] - Set \( y = 0 \): \[ 2x = 26 \implies x = 13 \] - The intercepts are \( (0, \frac{26}{3}) \) and \( (13, 0) \). 2. **For the inequality \( x + 2y \leq 16 \)**: - Find the boundary line: \[ x + 2y = 16 \] - To find the intercepts: - Set \( x = 0 \): \[ 2y = 16 \implies y = 8 \] - Set \( y = 0 \): \[ x = 16 \] - The intercepts are \( (0, 8) \) and \( (16, 0) \). ### Step 2: Determine the Feasible Region - The inequalities \( x \geq 0 \) and \( y \geq 0 \) restrict the region to the first quadrant. - The region satisfying \( 2x + 3y \leq 26 \) is below the line \( 2x + 3y = 26 \). - The region satisfying \( x + 2y \leq 16 \) is below the line \( x + 2y = 16 \). ### Step 3: Find the Intersection Points To find the intersection of the two lines, solve the system of equations: 1. \( 2x + 3y = 26 \) 2. \( x + 2y = 16 \) Let's solve this system. 1. From the second equation, express \( x \) in terms of \( y \): \[ x = 16 - 2y \] 2. Substitute into the first equation: \[ 2(16 - 2y) + 3y = 26 \] \[ 32 - 4y + 3y = 26 \] \[ 32 - y = 26 \implies y = 6 \] 3. Substitute \( y = 6 \) back into \( x = 16 - 2y \): \[ x = 16 - 2(6) = 4 \] The intersection point is \( (4, 6) \). ### Step 4: Graph the Region - Plot the lines and shade the regions that satisfy the inequalities. - The feasible region \( R \) is bounded by the lines and the axes, including the intersection point \( (4, 6) \). ### Conclusion The region \( R \) is the area in the first quadrant that is below both lines and includes the intersection point \( (4, 6) \). You can graph this using graphing software or by hand based on the steps provided.