Find the maximum value and the minimum value of the function and the values of $x$ and $y$ for which they occur. \[ P=16 x-5 y+68, \text { subject to } 5 x+9 y \leq 45,0 \leq y \leq 4, \text { and } 0 \leq x \leq 7 \]
Solution
Step 1 ：First, we need to find the vertices of the polygon formed by the constraints. The constraints are \(5x + 9y \leq 45\), \(0 \leq y \leq 4\), and \(0 \leq x \leq 7\). By solving these equations, we get the vertices as \({x: 9, y: 0}\), \({x: 9/5, y: 4}\), \({x: 0, y: 5}\), \({x: 7, y: 10/9}\), \({x: 0, y: 0}\), \({x: 7, y: 0}\), \({x: 0, y: 4}\), and \({x: 7, y: 4}\).
Step 2 ：Next, we substitute these vertices into the function \(P = 16x - 5y + 68\) to find the maximum and minimum values. The values we get are \(({x: 9, y: 0}, 212)\), \(({x: 9/5, y: 4}, 384/5)\), \(({x: 0, y: 5}, 43)\), \(({x: 7, y: 10/9}, 1570/9)\), \(({x: 0, y: 0}, 68)\), \(({x: 7, y: 0}, 180)\), \(({x: 0, y: 4}, 48)\), and \(({x: 7, y: 4}, 160)\).
Step 3 ：From these values, we can see that the maximum value of the function is 212 and it occurs at \(x=9\), \(y=0\). The minimum value of the function is 43 and it occurs at \(x=0\), \(y=5\).
Step 4 ：So, the maximum and minimum values of the function are \(\boxed{212}\) and \(\boxed{43}\) respectively.
