Maximize and minimize $p = 3 x - y$ subject to $$\begin{array} { l }
x + y \geq 3 \
x - y \leq 3 \
x - y \geq - 3 \
x \leq 12 , y \leq 12
\end{array}$$
A)Max: $p = 15$ , $x = 9$ , $y = 12$ , min: $p = - 3$ , $x = 0$ , $y = 3$
B) Max: $p = 27$ , $x = 12$ , $y = 9$ ,
Min: $p = 3$ , $x = 0$ , $y = 3$
C) Max: $p = 27$ , $x = 12$ , $y = 9$ ,
Min: $p = - 3$ , $x = 0$ , $y = 3$
D) the feasible region is empty
E) the objective function is unbounded

Question

Maximize and minimize $p = 3 x - y$ subject to   $$\begin{array} { l } 
x + y \geq 3 \
x - y \leq 3 \
x - y \geq - 3 \
x \leq 12 , y \leq 12
\end{array}$$  
A)Max: $p = 15$ , $x = 9$ , $y = 12$ , min: $p = - 3$ , $x = 0$ , $y = 3$
B) Max: $p = 27$ , $x = 12$ , $y = 9$ ,
Min: $p = 3$ , $x = 0$ , $y = 3$
C) Max: $p = 27$ , $x = 12$ , $y = 9$ ,
Min: $p = - 3$ , $x = 0$ , $y = 3$
D) the feasible region is empty
E) the objective function is unbounded

Quizplus · Accepted Answer

To find the maximum and minimum of $p = 3x - y$, we need to consider the constraints and identify the feasible region. The constraints are:1. $x + y \geq 3$2. $x - y \leq 3$3. $x - y \geq -3$4. $x \leq 12$, $y \leq 12$The feasible region is defined by the intersection of these constraints. To maximize or minimize $p$, we look at the vertices of the feasible region, which are determined by the intersection points of the constraints.For the maximum, setting $x = 12$ and adjusting $y$ to satisfy the constraints, we find that $y = 9$ gives a maximum value for $p$ since $3(12) - 9 = 36 - 9 = 27$.For the minimum, setting $x = 0$ and $y = 3$ satisfies all the constraints and gives $p = 3(0) - 3 = -3$.Therefore, the maximum value of $p$ is 27 when $x = 12$ and $y = 9$, and the minimum value of $p$ is -3 when $x = 0$ and $y = 3$.

Maximize and Minimize p=3x−yp = 3 x - yp=3x−y Subject To x+y≥3x−y≤3x−y≥−3x≤12,y≤12\begin{array} { l } x + y \geq 3 \\x - y \leq 3 \\x - y \geq - 3 \\x \leq 12 , y \leq 12\end{array}x+y≥3x−y≤3x−y≥−3x≤12,y≤12​

Maximize and Minimize $p = 3 x - y$ Subject To $\begin{array} { l } x + y \geq 3 \\x - y \leq 3 \\x - y \geq - 3 \\x \leq 12 , y \leq 12\end{array}$