There are no feasible solutions for minimum and maximum values for the function z = 2x + 4y that follow the given constraints.
- The given constraints are.
- 4x + y ≤ 40
- 20x +y ≥72
- 4x + 5y ≥ 72
- We change these inequalities to equations.
- 4x + y = 40
- 20x +y =72
- 4x + 5y = 72.
- Now we find the points which satisfy the equation when x is 0 and when y is 0. We do this for every equation.
- (0, 40) and (10, 0) satisfy the first equation.
- (0, 72) and (3.6, 0) satisfy the second equation.
- (0, 14.4) and (18, 0) satisfy the third equation.
- Now we plot these equations on the graph.
- Now we shade the regions that belong to the corresponding inequalities.
- The common region contains our feasible solution.
- But here we have no common region for all three inequalities.
- So, there are no feasible solutions.
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Answer:
286 sweets.
Step-by-step explanation:
If Adrian had x sweets, Charlie got x+52 sweets.
The number of parts is 7+4+11 = 22.
So Adrian had 7/22 of the total and Charlie had 11/22 = 1/2 of the total.
This means that Ben had 4/22 = 2/11 of the total.
If y is the total number of sweets:
y = x + x + 52 + 2y/11
9y/11 = 2x + 52
2x = 9y/11 - 52.............(1)
Also x + 52 = y/2
y = 2x + 104
2x = y - 104...............(2)
Equating (1) and (2) - (as there is 2x on left side of both equations):-
9y/11 - 52 = y - 104
52 = 2y/11
y = 52 * 11/2
y = 286 sweets.
Answer:
X = -3
Step-by-step explanation:
-5 and -8 are like terms, so you can add 5 to the other side. This leaves you with x = -3