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)

if the eqn of the parabola is f(x) = (x-4)(x+2), we need to mult. this out to find the coordinates of the vertex.

then f(x) = x^2 + 2x - 4x - 8, and f(x) = x^2 - 2x - 8.

one way in which you can find the vertex is to calculate x = -b / (2a).   here, that comes to

        )

x = = 1.   f(1) = 1^2 - 2(1) - 8 = - 9.   vertex is at (1, -9).

        2(1) 

find another point on the curve:   let x=0, for which y=-8.   then (0, -8) is on the curve.   plot the vertex and this point and draw a smooth parabola thru both.

hope this

i can answer this question sweetie!

step-by-step explanation: