Question

what is the minimum value of $c = 7x + 8y$, given the constraints on $x$ and $y$ listed below?
$2x + y \geq 8$
$x + y \geq 6$
$x \geq 0$
$y \geq 0$
\bigcirc 32
\bigcirc 42
\bigcirc 46
\bigcirc 64

Explanation:

Step1: Identify feasible region vertices

First, find the intersection points of the constraint lines and axes:

1. Intersection of $2x+y=8$ and $x+y=6$:

Subtract the second equation from the first: $(2x+y)-(x+y)=8-6$ → $x=2$. Substitute $x=2$ into $x+y=6$: $2+y=6$ → $y=4$. So point $(2,4)$.

1. Intersection of $2x+y=8$ and $x=0$: $y=8$, point $(0,8)$.
2. Intersection of $x+y=6$ and $y=0$: $x=6$, point $(6,0)$.
3. Intersection of $x=0$ and $y=0$ does not satisfy $2x+y\geq8$, so it is not in the feasible region.

Step2: Calculate C at each vertex

• For $(2,4)$: $C=7(2)+8(4)=14+32=46$
• For $(0,8)$: $C=7(0)+8(8)=0+64=64$
• For $(6,0)$: $C=7(6)+8(0)=42+0=42$

Step3: Find minimum C

Compare the calculated values: $42<46<64$

Answer:

42