C= 8x+6y
subject to constraints:
3x+2y > 160
5x+2y > 200
x+2y >80
x>0
y>0
This question is on graphing and shadowing the feasible region. I found all 4 points A=(80,0) B=(40,20) C=(20,50) D=(0,100) which are correct answers. The only thing i cant find is the isocost lines. The answer are C=400 and C=600. I cant draw or show my graphs here but anyone have idea how can i manually find the isocosts without graphing??
To manually find the isocost lines without graphing, you can use the given equation of the cost function, C = 8x + 6y, and the given values for C (400 and 600) to solve for x and y.
1. For C = 400:
Substitute C = 400 into the cost equation:
400 = 8x + 6y
From the given constraints, you can isolate one variable in terms of the other. Let's isolate x:
3x + 2y > 160 ⇒ x > (160 - 2y) / 3
Substitute the expression for x into the cost equation:
400 = 8((160 - 2y) / 3) + 6y
Simplify and solve for y:
400 = (1280 - 16y) / 3 + 6y
Multiply both sides by 3 to eliminate the fraction:
1200 = 1280 - 16y + 18y
1200 = 1280 + 2y
2y = 1200 - 1280
2y = -80
y = -40
Now substitute the value of y back into the constraint to find x:
x = (160 - 2(-40)) / 3
x = (160 + 80) / 3
x = 240 / 3
x = 80
So, for C = 400, you get the point (x, y) = (80, -40).
2. For C = 600:
Repeat the same process as above, but substitute C = 600 into the cost equation:
600 = 8x + 6y
Solve for y:
600 = 8((160 - 2y) / 3) + 6y
1800 = 1280 - 16y + 18y
1800 = 1280 + 2y
2y = 520
y = 260
Substitute the value of y back into the constraint to find x:
x = (160 - 2(260)) / 3
x = (160 - 520) / 3
x = (-360) / 3
x = -120
So, for C = 600, the point is (x, y) = (-120, 260).
These points (80, -40) and (-120, 260) represent the intersection of the isocost lines with the feasible region. Note that the negative y-value for C=400 suggests that the feasible region does not extend to negative values of y, so it is likely an error. Double-check your calculations to ensure accuracy.