Question

what is the maximum value of $p = 4x + 2y$, given the constraints on $x$ and $y$ listed below?
$x + 2y \leq 10$
$y \leq 2$
$x \geq 0$
$y \geq 0$
\bigcirc\\ 10
\bigcirc\\ 20
\bigcirc\\ 24
\bigcirc\\ 40

Explanation:

Step1: Identify feasible region vertices

Find intersection points of constraints:

1. $x=0, y=0$: $(0,0)$
2. $x=0, y=2$: $(0,2)$
3. $y=2, x+2y=10$: Substitute $y=2$ into $x+2y=10$, get $x=10-4=6$, so $(6,2)$
4. $y=0, x+2y=10$: $(10,0)$

Step2: Evaluate $P=4x+2y$ at vertices

• At $(0,0)$: $P=4(0)+2(0)=0$
• At $(0,2)$: $P=4(0)+2(2)=4$
• At $(6,2)$: $P=4(6)+2(2)=24+4=28$
• At $(10,0)$: $P=4(10)+2(0)=40$

Step3: Compare values to find maximum

The values are $0, 4, 28, 40$. The largest is $40$.

Answer:

40