Find the minimum and maximum values of the objective function, and the points at which these values occur subject to the given constraints. Round your answers to the nearest hundredth. Objective Function \[ \begin{array}{r} f(x, y)=3 x+15 y \quad x \geq 0 ; y \geq 0 \\ 12 x+6 y \leq 144 \\ 12 x+6 y \leq 288 \end{array} \]
Solution
Step 1 ：First, we need to find the vertices of the feasible region. The feasible region is defined by the constraints \(x \geq 0\), \(y \geq 0\), \(12x + 6y \leq 144\), and \(12x + 6y \leq 288\). The vertices are the points where these constraints intersect. In this case, the vertices are \((0, 0)\), \((0, 24)\), and \((12, 0)\).
Step 2 ：Next, we substitute these vertices into the objective function \(f(x, y) = 3x + 15y\). The values of the objective function at these points are 0, 360, and 36 respectively.
Step 3 ：We then find the minimum and maximum values of the objective function. The minimum value is 0 and the maximum value is 360.
Step 4 ：We also need to find the points at which these minimum and maximum values occur. The minimum value 0 occurs at the point \((0, 0)\) and the maximum value 360 occurs at the point \((0, 24)\).
Step 5 ：Finally, we can conclude that the minimum value of the objective function is 0 at the point \((0, 0)\) and the maximum value is 360 at the point \((0, 24)\). So, the answers are \(\boxed{0}\), \(\boxed{(0, 0)}\), \(\boxed{360}\), and \(\boxed{(0, 24)}\).
