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Oxana [17]
4 years ago
11

The UMass IIE student chapter is going to run a fundraising event from 1pm through 5pm. Due to the large scale of the event, stu

dent workers will be hired to work for the event and each worker will be assigned to a particular shift. There are three shifts: 1pm-3pm, 2pm-4pm, 3pm-5pm. The pay for each of the three shifts is $20, $24 and $28 per worker respectively. The minimum number of workers needed for each hour is listed below: 1pm-2pm: 20, 2pm-3pm: 25, 3pm-4pm: 25, 4pm-5pm: 30. ASCE would like to minimize the hiring cost while satisfying the demand of workers in each hour.
A) Define the decision variables.
B) Write down the objective function.
C) Write down the constraint that the demand of workers for 2pm-3pm is satisfied.
Mathematics
1 answer:
vagabundo [1.1K]4 years ago
4 0

Answer:

a. x1,x2,x3,x4

b. 20x1 + 24x2 + 28x3

c. x1 +x2 ≥ 25

Step-by-step explanation:

At a time that spans from 1 to 5pm, we have timing as follows:

1pm - 2pm = x1

2pm - 3pm = x2

3pm - 4pm = x3

4pm - 5pm = x4

a. The decision variables are x1, x2, x3, x4 and these are the number of people for each shift.

b. The objective function is the minimum cost which is

20x1 + 24x2 + 28x3

c. The constraints for each has been listed

1 - 2pm, x1 ≥ 20 workers

2 - 3pm, x1 + x2 ≥ 25

3 - 4pm, x2 + x3 ≥ 25

4 - 5pm, x3 ≥ 30

The question asked us to write the one for 2-3pm, which is x1 + x2 ≥ 25

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Factorise the expression: <br> x^2 - 144 <br><br> and<br><br> 2ax – 6ay + bx – 3by
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Answer:

(x-12)(x+12) and (2a+b)(x-3y)

Step-by-step explanation:

So first we have:

x^2-144

The key here is that there is no bx in the ax^2+bx+c equation here.

This means that b must equal 0, or in other words, cancel itself out.

To do this, first, the two factors must have opposite signs, and secondly, the blank values must be equal (x+_) * (x-_)

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Lets seperate the two parts of this equation:

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Now lets look at:

2ax - 6ay

How can we simplify this? Factor the 2a, which both have in common:

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Now lets look at and factor it too:

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Both of these have x-3y. Concidence? Nope!

From the way we split the equation into two pieces and factored, the two x-3ys are actually the same, and we can combine them!

Then, the 2a and the +b can be combined to give us:

(2a+b)

Puttting these together gives us:

(2a+b)(x-3y)

Hope this helps!

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