Question

Helpp

Answer 1

\[-x+3 \ge y\]
\[y \le \frac{ 1 }{ 3 }x + 1\]

let's rearrange in the form ax+by <= c

so we get

\[x+y \le 3\]
\[\frac{ -1 }{ 3 }x + y \le 1\]

and we want to maximize C=5x-4y

so lets draw our lines on a graph so for

\[x+y \le 3\]
when x=0, y=3
y=0, x=3

so draw a straight line from 3 to 3, from y axis to x axis

for

\[\frac{ -1 }{ 3 }x + y \le 1\]
we get
x=0, y=1
y=0, x=-3

draw it on the same graph



now from the objective function c=5x-4y
choose any number, i would recommend a multiple of 5x4 so 20
let c=20

x=0, y=-5
y=0, x=4
this is your gradient line that you will move up your graph, since we are maximizing, the last intersection will be your answer


increase the value of your objective function in multiples of 20
so lets try 40
so 40=5x-4y
x=0, y=-10
y=0, x=8


so we can see it got closer and closer to the intersection in the bottom right corner which was the last intersection it crossed
hence your answer is that vertice

so your answer is

y=0, x=3

This is from memory, but this is the best that I can help

Answer 2

Thank you so much!!