91Ó°ÊÓ

First, we graph the constraints to identify the feasible region of the LP: 1. \(5x + 2y \geq 40\) 2. \(x + 2y \geq 20\) 3. \(y \geq 3\) 4. \(x \geq 0\) To graph these inequalities, we can rewrite them as equalities and then find where the feasible region lies: 1. \(y = (40 - 5x)/2\) 2. \(y = (20 - x)/2\) 3. \(y = 3\) Plot these lines, while considering the last constraint \(x \geq 0\), to define the feasible region.

The feasible region is a polygon, and we can find the corner points by calculating the points where the constraints intersect. By inspection of the graph, the constraints intersect at the following points: 1. Intersection of \(5x + 2y = 40\) and \(x + 2y = 20\) 2. Intersection of \(x + 2y = 20\) and \(y = 3\) 3. Intersection of \(5x + 2y = 40\) and \(y = 3\) The last constraint \(x \geq 0\) will help us eliminate any points outside the feasible region.

We now calculate the value of the cost function at each corner point: 1. Intersection of \(5x + 2y = 40\) and \(x + 2y = 20\): Solve the linear system of equations: \(5x + 2y = 40\) \(x + 2y = 20\) From the second equation, \(y = 10 - \frac{1}{2}x\). Substituting this into the first equation, we get \(5x + 2(10 - \frac{1}{2}x) = 40\). Solving for x, we get \(x = 8\). Plugging this value back into the second equation, we find \(y = -2\). Since \(y < 3\), this point is not in the feasible region and will not be considered. 2. Intersection of \(x + 2y = 20\) and \(y = 3\): Substitute \(y = 3\) into the equation \(x + 2y = 20\), we get \(x = 14\). Thus, the point is (14, 3). Calculate the cost function value: \(C = 2(14) + 10(3) = 28 + 30 = 58\). 3. Intersection of \(5x + 2y = 40\) and \(y = 3\): Substitute \(y = 3\) into the equation \(5x + 2y = 40\), we get \(x = 4\). Thus, the point is (4, 3). Calculate the cost function value: \(C = 2(4) + 10(3) = 8 + 30 = 38\).

The feasible corner points are (14, 3) and (4, 3), with cost function values of 58 and 38, respectively. Since we want to minimize the cost function, the optimal solution is (4, 3) with a cost of C = 38.