Find the maximum value and the minimum value of the function and the values of $x$ and $y$ for which they occur. \[ P=15 x-4 y+69, \text { subject to } 7 x+8 y \leq 56,0 \leq y \leq 4 \text {, and } 0 \leq x \leq 6 \] The maximum value $\square$ occurs where $x=\square$ and $y=\square$. The minimum value $\square$ occurs where $x=\square$ and $y=\square$. (Do not round until the final answer. Then round to two decimal places as needed.)
Solution
Step 1 ：Identify the feasible region by graphing the constraints. The inequality \(7x+8y\leq56\) can be rewritten as \(y\leq7-\frac{7}{8}x\). This is a line with a negative slope, and the feasible region is below this line. The inequalities \(0\leq y\leq4\) and \(0\leq x\leq6\) define a rectangle in the first quadrant. The feasible region is the intersection of the area below the line and the rectangle.
Step 2 ：Find the vertices of the feasible region. These are the points where the constraints intersect. These points are \((0,0)\), \((0,4)\), \((6,0)\), and the point where the line \(y=7-\frac{7}{8}x\) intersects the line \(x=6\). To find this point, substitute \(x=6\) into the equation of the line to get \(y=7-\frac{7}{8}*6=7-\frac{21}{4}=7-5.25=1.75\). So the fourth vertex is \((6,1.75)\).
Step 3 ：Evaluate the function at the vertices. Substitute the coordinates of the vertices into the function \(P=15x-4y+69\) to find the maximum and minimum values. At \((0,0)\), \(P=15*0-4*0+69=69\). At \((0,4)\), \(P=15*0-4*4+69=69-16=53\). At \((6,0)\), \(P=15*6-4*0+69=90+69=159\). At \((6,1.75)\), \(P=15*6-4*1.75+69=90-7+69=152\).
Step 4 ：Identify the maximum and minimum values. The maximum value of the function is \(159\), which occurs at \(x=6\) and \(y=0\). The minimum value of the function is \(53\), which occurs at \(x=0\) and \(y=4\). So, the maximum value \(\boxed{159}\) occurs where \(x=6\) and \(y=0\). The minimum value \(\boxed{53}\) occurs where \(x=0\) and \(y=4\).
