Maximize $p = x + 2 y$
Subject to
$$\begin{aligned}
x + y & \leq 20 \
y & \geq 8 \
2 x - y & \geq 0 \
x \geq 0 , y & \geq 0
\end{aligned}$$
A) $p = 80$ , $x = 40$ , $y = 20$
B) $p = 100$ , $x = 20$ , $y = 40$
C) $p = \frac { 80 } { 3 }$ , $x = \frac { 40 } { 3 }$ , $y = \frac { 20 } { 3 }$
D) $p = \frac { 100 } { 3 }$ , $x = \frac { 20 } { 3 }$ , $y = \frac { 40 } { 3 }$
E) no solution

Question

Maximize   $p = x + 2 y$  
Subject to
  $$\begin{aligned}
x + y & \leq 20 \
y & \geq 8 \
2 x - y & \geq 0 \
x \geq 0 , y & \geq 0
\end{aligned}$$  
A) $p = 80$ , $x = 40$ , $y = 20$
B) $p = 100$ , $x = 20$ , $y = 40$
C) $p = \frac { 80 } { 3 }$ , $x = \frac { 40 } { 3 }$ , $y = \frac { 20 } { 3 }$
D) $p = \frac { 100 } { 3 }$ , $x = \frac { 20 } { 3 }$ , $y = \frac { 40 } { 3 }$
E) no solution

Quizplus · Accepted Answer

The feasible region is bounded by the constraints. The optimal solution occurs at the intersection of $x + y = 20$ and $2x - y = 0$, which gives $x = \frac{20}{3}$ and $y = \frac{40}{3}$. Substituting these into $p = x + 2y$ gives $p = \frac{100}{3}$.

Maximize p=x+2yp = x + 2 yp=x+2y Subject To x+y≤20y≥82x−y≥0x≥0,y≥0\begin{aligned}x + y & \leq 20 \\y & \geq 8 \\2 x - y & \geq 0 \\x \geq 0 , y & \geq 0\end{aligned}x+yy2x−yx≥0,y​≤20≥8≥0≥0​

Maximize $p = x + 2 y$ Subject To $\begin{aligned}x + y & \leq 20 \\y & \geq 8 \\2 x - y & \geq 0 \\x \geq 0 , y & \geq 0\end{aligned}$