Question

16 - 20: show your work: solve the system of inequalities and name the vertices of the polygonal convex set. then, find the maximum and minimum value of f(x,y).
y ≥ 0
x ≤ 4
x ≥ 0
4y - 2x ≤ 8
f(x,y)=3x - 5y

Explanation:

Step1: Rewrite the inequality

Rewrite $4y - 2x\leq8$ as $y\leq\frac{1}{2}x + 2$.

Step2: Find the intersection points (vertices)

• Intersection of $x = 0$ and $y = 0$: Vertex is $(0,0)$.
• Intersection of $x = 0$ and $y=\frac{1}{2}x + 2$: Substitute $x = 0$ into $y=\frac{1}{2}x + 2$, we get $y = 2$. Vertex is $(0,2)$.
• Intersection of $x = 4$ and $y=\frac{1}{2}x + 2$: Substitute $x = 4$ into $y=\frac{1}{2}x + 2$, $y=\frac{1}{2}\times4+ 2=4$. Vertex is $(4,4)$.
• Intersection of $x = 4$ and $y = 0$: Vertex is $(4,0)$.

Step3: Evaluate the function $f(x,y)=3x - 5y$ at vertices

• At $(0,0)$: $f(0,0)=3\times0-5\times0 = 0$.
• At $(0,2)$: $f(0,2)=3\times0-5\times2=- 10$.
• At $(4,4)$: $f(4,4)=3\times4-5\times4=12 - 20=-8$.
• At $(4,0)$: $f(4,0)=3\times4-5\times0 = 12$.

Answer:

Vertices: $(0,0),(0,2),(4,4),(4,0)$; Maximum value of $f(x,y)$ is $12$ at $(4,0)$; Minimum value of $f(x,y)$ is $-10$ at $(0,2)$