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Reil [10]
3 years ago
5

Rachel mows 5 lawns in 8 hours. At this rate how many lawns can she mow in 40 hour?

Mathematics
1 answer:
posledela3 years ago
5 0

We can think of this as an equality

5 Lawns = 8 Hours

If she needs 40 hours, you can multiply 8 hours by 5 to get 40, but now it's not balanced. So, you need to multiply 5 to the other side too so:

5 x (5) = 8 x (5)

Therefore she mows 25 Lawns in 40 Hours

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A marketing research firm would like to survey undergraduate and graduate college students about whether or not they take out st
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Answer:

Objective Minimize 10x1 +15x2 + 12x3+18x4+15x5+ 21x6

The central total cost $ 17586 due to the number of central undergraduate students 1458 is very high.

The total minimum cost would $ 17586 +$260+ $300= $ 18146

Step-by-step explanation:

Let

X1 = # of undergraduate students from the East region,

X2 = # of graduate students from the East region,

X3 = # of undergraduate students from the Central region,

X4 = # of graduate students from the Central region,

X5 = # of undergraduate students from the West region, and

X6 = # of graduate students from the West region.

Then  the cost functions are

y1= 10x1 +15x2

y2= 12x3+18x4

y3= 15x5+ 21x6

According to the given conditions

The constraints are

x1 +x2 + x3+x4+ x5+ x6 ≥ 1500------- A

15X2 +18X4+21X6 ≥ 400---------B

21X6 ≥ 100

X6 ≥ 100/21

X6 ≥ 4.76

Taking

X6= 5

10X1 ≤ 500

X1 ≤ 500/10

X1≤ 5

18X4 ≥ 75

X4 ≥ 75/18

X4 ≥ 4.167

Taking

X4= 5

Putting the values

15X5+ 21X6 ≥ 300

15X5+ 21(5) ≥ 300

15X5+ 105 ≥ 300

15X5 ≥ 300-105

15X5 ≥ 195

X5 ≥ 195/15

X5 ≥ 13

Putting value of X6 and X4 in B

15X2 +18X4+21X6 ≥ 400

15X2 +18(5)+21(5) ≥ 400

15X2 +195 ≥ 400

15X2  ≥ 400-195

15X2  ≥ 205

X2  ≥ 205/15

X2  ≥ 13.67

Taking X2= 14

Now putting the values in the cost equations to check whether the conditions are satisfied.

y1= 10x1 +15x2

y1= 10 (5) + 15(14)= 50 + 210= $ 260

y3= 15x5+ 21x6

y3= 15 (13) + 21(5)

y3= 195+105= $ 300

x1 +x2 + x3+x4+ x5+ x6 ≥ 1500

5+14+x3+5+13+5≥ 1500

x3≥ 1500-42

x3≥ 1458

y2= 12x3+18x4

y2= 12 (1458)  + 18 (5)

y2= 17496 +90

y2= $ 17586

The cost can be minimized if the number of students from

                                 Undergraduate         Graduate

East Region              X1≤ 5                            X2  ≥ 13.67

Central                      X3≥ 1458                   X4 ≥ 4.167

West                          X5 ≥ 13                          X6 ≥ 4.76

This will result in the required number of students that is 1500

Constraints:

East Undergraduate must not be greater than 5

East Graduate must not be less than 13

Central Undergraduate must  be greater than 1458

Central Graduate must  be greater than 4

West Undergraduate must  be greater than 13

West Graduate must  be greater than 4

The central total cost $ 17586 due to the number of central undergraduate students 1458 is very high.

The east region has a least cost of $260 and west region has a cost of $300.

The total minimum cost would $ 17586 +$260+ $300= $ 18146

6 0
3 years ago
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