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Maurinko [17]
3 years ago
11

Find the maximum value of P=3x+2y subject to the following constraints: Please help me!!!

Mathematics
1 answer:
UkoKoshka [18]3 years ago
6 0

Answer:

maximum value of P is 33

Step-by-step explanation:

Sketch the lines represented by the constraints

5x + y = 16

with intercepts (\frac{16}{5}, 0) and (0, 16)

2x + 3y = 22

with intercepts (11, 0) and (0, \frac{22}{3})

The solutions to both are below the lines.

Solve 5x + y = 16 and 2x + 3y = 22 simultaneously to find the point of intersection at (2, 6)

The coordinates of the vertices of the feasible region are

(0, 0), (0, 16), (2, 6), 11, 0)

Evaluate the objective function at each vertex

P = 3(0) + 2(0) = 0 + 0 = 0

P = 3(0) + 2(16) = 0 + 32 = 32

P = 3(2) + 2(6) = 6 + 12 = 18

P = 3(11) + 2(0) = 33 + 0 = 33

The maximum value of P is 33 when x = 11 and y = 0

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The variable Z is directly proportional to X. When X is 12, Z has the value 228. What is the value of Z when X = 18
Kisachek [45]
Here, z = kx

k = z/x
k = 228/12
k = 19

So, when x = 18, 
z = 19 * 18
z = 342

In short, Your Answer would be: 342

Hope this helps!
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3 years ago
5s + 7s????????????????
natali 33 [55]

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12s

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This is your answer

8 0
3 years ago
Read 2 more answers
A sphere rolls up an inclined plane of inclination angle 30 degree. At the bottom of the incline the center of mass of the spher
Zigmanuir [339]

Answer: The sphere is 0.002548 cm travel up the plane.

Step-by-step explanation:

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As we know that

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and

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l=\dfrac{2Mr^2}{5}

So, it becomes,

KE=0.5\times \dfrac{2Mr^2}{5}w^2=Mgd\sin \theta

And 0.25\ m/sv=rw

So, it becomes,

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Hence, the sphere is 0.002548 cm travel up the plane.

5 0
3 years ago
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