Maximize $p = 12 x + 10 y$
Subject to
$$\begin{aligned}
x + y & \leq 40 \
x & \geq 1 \
- x + 2 y & \geq 0 \
x \geq 0 , y & \geq 0
\end{aligned}$$
A) $p = 1,360$ , $x = 80$ , $y = 40$
B) $p = \frac { 560 } { 3 }$ , $x = \frac { 40 } { 3 }$ , $y = \frac { 80 } { 3 }$
C) $p = \frac { 1,360 } { 3 } , x = \frac { 80 } { 3 } , y = \frac { 40 } { 3 }$
D) $p = 1,280$ , $x = 40$ , $y = 80$
E) no solution

Question

Maximize $p = 12 x + 10 y$  
Subject to
  $$\begin{aligned}
x + y & \leq 40 \
x & \geq 1 \
- x + 2 y & \geq 0 \
x \geq 0 , y & \geq 0
\end{aligned}$$ 
A) $p = 1,360$ , $x = 80$ , $y = 40$
B)  $p = \frac { 560 } { 3 }$ , $x = \frac { 40 } { 3 }$ , $y = \frac { 80 } { 3 }$
C) $p = \frac { 1,360 } { 3 } , x = \frac { 80 } { 3 } , y = \frac { 40 } { 3 }$
D) $p = 1,280$ , $x = 40$ , $y = 80$
E) no solution

Quizplus · Accepted Answer

The constraints form a feasible region on the coordinate plane. The objective function $p = 12x + 10y$ needs to be maximized. By evaluating the objective function at the vertices of the feasible region created by the constraints, it can be determined that option C gives the maximum value of $p$ that satisfies all the given constraints. The values in option C are plausible and respect the constraints, unlike the other options which either violate the constraints or do not provide the maximum value.

Maximize p=12x+10yp = 12 x + 10 yp=12x+10y Subject To x+y≤40x≥1−x+2y≥0x≥0,y≥0\begin{aligned}x + y & \leq 40 \\x & \geq 1 \\- x + 2 y & \geq 0 \\x \geq 0 , y & \geq 0\end{aligned}x+yx−x+2yx≥0,y​≤40≥1≥0≥0​

Maximize $p = 12 x + 10 y$ Subject To $\begin{aligned}x + y & \leq 40 \\x & \geq 1 \\- x + 2 y & \geq 0 \\x \geq 0 , y & \geq 0\end{aligned}$