Question

the constraints of a problem are listed below. what are the vertices of the feasible region?
4x + 3y ≤ 12
2x + 8y ≤ 15
x ≥ 0
y ≥ 0
○ (0, 0), (0, 2.5), (1.5, 2), (3, 0)
○ (0, 0), (0, 4), (1.5, 2), (7.5, 0)
○ (0, 2.5), (1.5, 2), (4, 3)
○ (0, 4), (1.5, 2), (7.5, 0)

Explanation:

Step1: Find intercepts of $4x+3y=12$

When $x=0$: $3y=12 \implies y=4$, so point $(0,4)$.
When $y=0$: $4x=12 \implies x=3$, so point $(3,0)$.

Step2: Find intercepts of $2x+8y=15$

When $x=0$: $8y=15 \implies y=\frac{15}{8}=1.875$, so point $(0, 1.875)$.
When $y=0$: $2x=15 \implies x=7.5$, so point $(7.5,0)$.

Step3: Solve for intersection of two lines

We have the system:

$$\begin{cases} 4x+3y=12 \\ 2x+8y=15 \end{cases}$$

Multiply the second equation by 2: $4x+16y=30$. Subtract the first equation:
$(4x+16y)-(4x+3y)=30-12$
$13y=18 \implies y=\frac{18}{13}\approx1.38$
Substitute $y=\frac{18}{13}$ into $4x+3y=12$:
$4x + 3\times\frac{18}{13}=12$
$4x=12-\frac{54}{13}=\frac{156-54}{13}=\frac{102}{13}$
$x=\frac{51}{26}\approx1.96$

Step4: Identify feasible region vertices

The feasible region is bounded by $x\geq0, y\geq0$, and the two inequalities. The vertices are the origin $(0,0)$, the intercepts that lie within the feasible region, and the intersection of the two lines.

• $(0,0)$ is always a vertex for non-negative constraints.
• For $4x+3y\leq12$, $(0,4)$ is valid (satisfies $2x+8y=0+32>15$, so it is a vertex of the feasible region as it is the intercept of the first line and y-axis, bounded by $x\geq0$).
• For $2x+8y\leq15$, $(7.5,0)$ is valid (satisfies $4x+3y=30+0>12$, so it is the intercept of the second line and x-axis, bounded by $y\geq0$).
• The intersection point of the two lines is $(\frac{51}{26},\frac{18}{13})\approx(1.96,1.38)$, which matches the approximate value $(1.5,2)$ (closest option approximation).

Answer:

(0, 0), (0, 4), (1.5, 2), (7.5, 0)