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Paul [167]
3 years ago
13

What is the minimum value of the objective function, C with given constraints? C=5x+3y {⎨x+3y≤9 {5x+2y≤20 {x≥1 {y≥2

Mathematics
1 answer:
Solnce55 [7]3 years ago
3 0

Answer:

The answer is below

Step-by-step explanation:

Plotting the following constraints using the online geogebra graphing tool:

x + 3y ≤ 9         (1)

5x + 2y ≤ 20    (2)

x≥1 and y≥2      (3)

From the graph plot, the solution to the constraint is A(1, 2), B(1, 2.67) and C(3, 2).

We need to minimize the objective function C = 5x + 3y. Therefore:

At point A(1, 2): C = 5(1) + 3(2) = 11

At point B(1, 2.67): C = 5(1) + 3(2.67) = 13

At point C(3, 2): C = 5(3) + 3(2) = 21

Therefore the minimum value of the objective function C = 5x + 3y is at point A(1, 2) which gives a minimum value of 11.

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for the school play adult tickets cost 4$ and children tickets cost 2$ natalie is working at the ticket counter and just sold 20
lilavasa [31]
Let us formulate the independent equation that represents the problem. We let x be the cost for adult tickets and y be the cost for children tickets. All of the sales should equal to $20. Since each adult costs $4 and each child costs $2, the equation should be

4x + 2y = 20

There are two unknown but only one independent equation. We cannot solve an exact solution for this. One way to solve this is to state all the possibilities. Let's start by assigning values of x. The least value of x possible is 0. This is when no adults but only children bought the tickets.

When x=0,
4(0) + 2y = 20
y = 10

When x=1,
4(1) + 2y = 20
y = 8

When x=2,
4(2) + 2y = 20
y = 6

When x=3,
4(3) + 2y= 20
y = 4

When x = 4,
4(4) + 2y = 20
y = 2

When x = 5,
4(5) + 2y = 20
y = 0

When x = 6,
4(6) + 2y = 20
y = -2

A negative value for y is impossible. Therefore, the list of possible combination ends at x =5. To summarize, the combinations of adults and children tickets sold is tabulated below:

   Number of adult tickets             Number of children tickets
                  0                                                   10
                  1                                                    8
                  2                                                    6
                  3                                                    4
                  4                                                    2
                  5                                                    0




6 0
4 years ago
If a child's knowledge of the alphabet is limited to the letters a, b, c, i, and e, and if the child writes two letters at rando
Savatey [412]
The formula for probability is # of favorable outcomes divided by the total number of outcomes. 

So we know that there are a total of 5 letters. The child writes two letters, so we will have x/5 * x/5. Now, it is asking, what is the probability that they are both vowels.

We can see that a, i, and e are vowels, so the probability if the child were to write one letter would be 3/5. Since he is writing 2, you can multiply 3/5 * 3/5, which would give you your answer, 9/25. Therefore the probability that both letters are vowels is 9/25.

Hope this helps! Please rate, leave a thanks, and mark a brainliest answer. (Not necessarily mine). Thanks, it really helps! :D


7 0
3 years ago
Find the value of y such that the distance between the given points is indicated..... (5,y) and (8,-1) is 5.
maw [93]
(5,y)(8,-1)
d = sqrt ((x2 - x1)^2 + (y2 - y1)^2)
d = sqrt ((8 - 5)^2 + (-1 - y)^2)
d = sqrt (3^2 + (-1 - 3)^2
d = sqrt (9 + (-4^2)
d = sqrt (9 + 16)
d = sqrt 25
d = 5

so ur points are : (5,-3)(8,-1)
6 0
3 years ago
What is the answer to this problem 3z+4=34
finlep [7]
Solve for z

3z+4=34

Subtract 4 from both sides. 

= 3z=30

Divide both sides by 3

z=10


Hope I helped! :D

3 0
3 years ago
Read 2 more answers
Once again I am in need of help,
natka813 [3]
Hope it helps...........


3 0
4 years ago
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