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bekas [8.4K]
4 years ago
10

Find the minimum value of C=2x+y subject to the following constraints: Please help me!!

Mathematics
1 answer:
-BARSIC- [3]4 years ago
7 0

Answer:

The minimum value of C is 20

Step-by-step explanation:

we have

2x+3y\geq 24 ----> constraint A

4x+y\geq 38 ----> constraint B

x\geq 0 ----> constraint C

y\geq 0 ----> constraint D

Using a graphing tool

The solution set of the constraints is the shaded area

see the attached figure

The vertices of the shaded area are (0,38),(9,2) and (12,0)

To determine the minimum value of C substitute the values of x and y of each vertex in the objective function and then compare the results

C=2x+y

For (0,38) ---> C=2(0)+38=38

For (9,2) ---> C=2(9)+2=20

For (12,0) ---> C=2(12)+0=24

therefore

The minimum value of C is 20

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Answer:

x= 23

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Step-by-step explanation:

x-y=16

4y-5=x

4y-5-y=16

3y-5=16 add 5 to both sides

3y=21 divide by 3 on both sides

y=7

x-7=16 substitute back in

x=23 added 7 to both sides

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Step-by-step explanation:

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Kendra is considering two different long-distance phone plans. Phone plan A charges a $100 sign-up fee and 3 cents per minute. P
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Answer:

Part 1) The number of minutes in a month must be greater than 50 in order for the plan A to be preferable

Part 2) The number of minutes in a month must be equal to 50 minutes

Step-by-step explanation:

<u><em>The question is</em></u>

Part 1) How many minutes would Kendra have to use in a month in order for the plan A to be preferable? Round your answer to the nearest minute

Part 2) Enter the number of minutes where Kendra will pay the same amount for each long distance phone plan

Part 1)

Let

x ---> the number of minutes

we have

<em>Cost Plan A</em>

3x+100

<em>Cost Plan B</em>

5x

we know that

In order for plan A to be cheaper than plan B, the following inequality must hold true.

cost of plan A < cost of plan B

substitute

3x+100 < 5x

solve for x

subtract 3x both sides

100< 5x-3x\\100

divide by 2 both sides

50 < x

Rewrite

x> 50\ min

therefore

The number of minutes in a month must be greater than 50 in order for the plan A to be preferable

Part 2)

Let

x ---> the number of minutes

we have

<em>Cost Plan A</em>

3x+100

<em>Cost Plan B</em>

5x

we know that

In order for plan A cost the same than plan B, the following equation must hold true.

cost of plan A = cost of plan B

substitute

3x+100= 5x

solve for x

5x-3x=100\\2x=100\\x=50\ min

therefore

The number of minutes in a month must be equal to 50 minutes

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The answer would be true.
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