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Anna35 [415]
3 years ago
12

Linear programming subjected to constraints

Mathematics
1 answer:
anyanavicka [17]3 years ago
8 0

Answer:

The maximum value of C is 14

Step-by-step explanation:

we have the following constraints

x\geq 0 ----> constraint A

y\geq 0 ---> constraint B

2x+2y\leq 10 ---> constraint C        

3x+y\leq 9 ---> constraint D

Determine the area of the feasible region using a graphing tool

see the attached figure

The vertices of the feasible region are

(0,0),(0,5),(2,3),(3,0)

To find out the maximum value of the objective function C, substitute the value of x and the value of y of each vertices in the objective function an then compare the results

we have

C=4x+2y

For (0,0) ----> C=4(0)+2(0)=0

For (0,5) ----> C=4(0)+2(5)=10

For (2,3) ----> C=4(2)+2(3)=14

For (3,0) ----> C=4(3)+2(0)=12

therefore

The maximum value of C is 14

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Answer:

x=7

Step-by-step explanation:

Solve.

7x-5x-3=x+4

Calculate like terms.

7x-5x=2x

Rearrange.

2x-3=x+4

Apply inverse addition.

2x-3+3=x+4+3

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Isolate 2x by subtraction.

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7 0
3 years ago
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ValentinkaMS [17]
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Converting 30ft/s to m/s, initial velocity was 30ft/s x 0.305ft/meter = 9.14m/s 

Let's find how long it took for the velocity to equal zero, meaning when the ball reached it's highest point and, for a split second, stopped in mid-air before falling back down.

V(t) = Vi + a*t , where V(t) is velocity as a function of time, a is acceleration due to gravity, and t is time. Set V(t) = 0
0 = 9.14 + (-9.8)* t          Add -9.8t to both sides
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t = 0.93 seconds

Let's say your hand is the base point, or where h=zero. We want to find how high above your hand the ball went before it started coming down. Using the distance, or in this case height, formula:
h = Vi*t + (1/2)at²               Plug in Vi, a, and our t value, 0.93
h= 9.14 * 0.93 + (1/2)(9.8)(0.93²)
h= 8.5 + 4.9 (0.865)
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The ball made it 12.74 meters above your hand. Your friends had was one foot above yours, so let's subtract .305 meters to see how far it dropped from the peak height to his hand.
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Let's use the distance formula again to see how long it took to come down. Remember that this time, initial velocity is zero, since the ball starts off suspended in the air.
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2.5374 = t²
t = 1.59

The ball took .93 seconds to go up, and 1.59 seconds to come down to your friend's glove. The total time the ball was in the air:
.93 + 1.59 = 2.52 seconds
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Answer:

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Step-by-step explanation:

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