What point in the feasible region maximizes the objective function constraints: x> =0 y> =0 -x+3> =y y< =1/3 x+1 objective function: c=5x-4y explain!

So, the maximum of C is 15 at (3,0) and the minimum value is 0 at (0,0)

Step-by-step explanation:

When you graph all of the constraint functions, the vertices of the shape that is made are at the points above. Now you plug those points into the function equation C=5x-4y and see where you get a max or min.

(0,1)

(1.5,1.5)

(0,0)

(3,0)

(1,1)=C=1

(1.5,1.5)=C=1.5

(0,0)=C=0

(3,0)=C=15

So, the maximum of C is 15 at (3,0) and the minimum value is 0 at (0,0)

the correct answers i think are b. and d.

give me brainliest if it is correct!

region b will make the inequalities y < 2x + 1 and y > -x - 2 true ⇒ answer a

step-by-step explanation:

* lets study the graph and the inequality to solve the question

- the inequality y > -x - 2

∵ the form of the inequality is y > mx + c , m is the slope of the line

  and c is the y-intercept (the line intersect y-axis at point (0 , c)

∴ c = -2 that means the line intersect the y-axis at point (0 , -2)

∵ the sign of the inequality is >

∴ we shad the part over the line

  region a and b

- the inequality y < 2x + 1

∵ the form of the inequality is y > mx + c , m is the slope of the line

  and c is the y-intercept (the line intersect y-axis at point (0 , c)

∴ c = 1 that means the line intersect the y-axis at point (0 , 1)

∵ the sign of the inequality is <

∴ we shad the part under the line

  region b and c

∵ the common region between the two lines is region b

∴ region b will make the inequalities y < 2x + 1 and y > -x - 2 true