Question

help

Answer 1

i need help on my math problem

Answer 2

Then just write them out...You don't have to write a novel.

Answer 3

giving the system of constraints, name all the vertices. Then find the maximum value of the given objective function

Answer 4

???

Answer 5

hang on don't worry I will get the right answer for you

Answer 6

this is the problem

Answer 7

??? help meeee

Answer 8

with what

Answer 9

that question above

Answer 10

can i please get some help

Answer 11

You should start by graphing the feasible region. You already have that. Next you should state the possible points that could result in obtaining the max value of the objective function. You already have the possible points.

Lastly, you simply test the points algebraically by plugging in each point into the objective function. So here are the steps:

1. Draw the graph of the feasible region
2. Express the possible points (the intersections of the given lines)
3. Test each point by plugging them in one at a time into the objective function
4. State the max value and the point that resulted in producing the max value.

It's really not that difficult.

Answer 12

explain more

Answer 13

You have to graph this one out. Make all the lines- you'll end up using only the positive quadrant (that's why they gave you y>0 and x>0). When you graph all of the equations, don't pay attention to the less than or greater than signs until you're all done graphing. Once you're done, you'll get some sort of shape. To find the feasible region, shade in the shape using the greater than and less than signs. If the equation says greater than, shade above the line, but if it says less than, shade below it. The vertexes of the feasible region are just the corners of the shape that you just shaded in.
For the last part, to find maximum value, plug each vertex into your objective function. Like, if you were to get (5,3) THIS IS JUST AN EXAMPLE, you would put in c=6•5-4•3, and you'd get 18!!! Do this for each vertex. whichever number is the biggest will be your maximum value.
Hope this helped! :)

Answer 14

i have no calculator

Answer 15

second equation re-arranged, and first equation:
y =< - x + 5
y<=1/3x+3

these are inequality lines, of the 4 quadrants only the one
right up is in the feasable region because x,y>=0

Answer 16

first restriction of the feasible region
by y <= 1/3x +3:
1/3 x + 3
Reply Using Drawing


verteces: y -intercept, x- intercept of the function

Answer 17

just y - intercept and borders of the quadrant

Answer 18

the vertices for making the limitation on the left are
(0, 0) and (0, 3)

the vertex for the limitation on the right end is
(5, 0)

the vertex on the top of the object is the position
where the lines cross and make a shared point

Answer 19

so the last vertex is where these equations cross:
y <= -x +5
y <=1/3x +3

-x +5 = 1/3x +3 |+x
5 = 4/3 x +3 |-3
2 = 4/3 x |x3
6 = 4 x |4
x = 4/6

and to get y plug x into any equation?
y = -x + 5
y = -4/6 + 5
y = -4/6 + 30/6
y = 26/6

point:
(2/3, 4 1/3) if the algebra was correct..

Answer 20

maximum for c+4x-3y

Answer 21

c=4x+3y

Answer 22

https://www.desmos.com/calculator/ygtpnkixcp