(a) Shade the region, \( P \), in the \( x-y \) plane which satisfies simultaneously the inequalities: \( x+y \geq 5 \) \( 2 y-x \geq 0 \) \( x+5 y \leq 21 \) (b) Use the diagram in 4 (a) to find on the region, \( P \). the: L. Minimum value of \( x \), ii. Maximum value of ( \( 3 x+2 y \) ).

To tackle this problem, let’s first visualize the inequalities in the \(x-y\) plane: 1. For \(x + y \geq 5\): This is a line with a y-intercept of 5 and an x-intercept of 5. The shaded region will be above this line. 2. For \(2y - x \geq 0\) (or \(y \geq \frac{x}{2}\)): This line has a slope of \( \frac{1}{2} \), starting from the origin, and the shaded region will be above this line. 3. For \(x + 5y \leq 21\): This line can be rearranged to \(y \leq \frac{21 - x}{5}\), with a y-intercept of 4.2 and an x-intercept of 21. The shaded region will be below this line. Sketching these lines and identifying the overlapping area gives us the feasible region \(P\). Now, regarding the values requested in part (b): **i. Minimum value of \(x\):** By examining the points of intersection of the lines that bound the region \(P\), you can determine the vertices of the feasible region. The point where both inequalities meet at the lowest x-value (often occurs at the intersection of \(x + y = 5\) and \(y = \frac{x}{2}\)) typically gives the minimum x. Solve those equations simultaneously to find that minimum. **ii. Maximum value of \(3x + 2y\):** After identifying the vertices of the feasible region from the intersections, you will evaluate the expression \(3x + 2y\) at each vertex. The largest value from these calculations will determine the maximum value of \(3x+2y\), which lies at a corner point of the region \(P\). Use a systematic approach with substitution to find which vertex yields the maximum. Happy graphing!