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allochka39001 [22]
3 years ago
5

For the feasibility region shown below, find the maximum value of the function P=3x+2y.

Mathematics
2 answers:
Setler [38]3 years ago
6 0

Answer:

28

Step-by-step explanation:


abruzzese [7]3 years ago
5 0
Given:
P = 3x + 2y

There is an accompanying graph in this problem. In the graph, there are 4 points to consider. I'll just assign letters on each point.

Point O is found in x = 0 ; y = 0 or (0,0)
Point A is found in x = 8 ; y = 0 or (8,0)
Point B is found in x = 6 ; y = 5 or (6,5)
Point C is found in x = 0 ; y = 8 or (0,8)

We will substitute x and y in the equation by its values per point.
Point A = 3(8) + 2(0) = 24 + 0 = 24
Point B = 3(6) + 2(5) = 18 + 10 = 28
Point C = 3(0) + 2(8) = 0 + 16 = 16 

The maximum value of the function P = 3x+2y is 28 and its minimum value is 16.
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Answer:

1 roll of ribbon, 1 package of buttons, and 3 packages of beads

Step-by-step explanation:

Use the information you found about how much each makes to answer the question.

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3 years ago
If s is an increasing function, and t is a decreasing function, find Cs(X),t(Y ) in terms of CX,Y .
Sedbober [7]

Answer:

C(X,Y)(a,b)=1−C(s(X),t(Y))(a,1−b).

Step-by-step explanation:

Let's introduce the cumulative distribution of (X,Y), X and Y :

F(X,Y)(x,y)=P(X≤x,Y≤y)

  • FX(x)=P(X≤x)
  • FY(y)=P(Y≤y).

Likewise for (s(X),t(Y)), s(X) and t(Y) :

F(s(X),t(Y))(u,v)=P(s(X)≤u

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  • Ft(Y)(v)=P(t(Y)≤v).

Now, First establish the relationship between F(X,Y) and F(s(X),t(Y)) :

F(X,Y)(x,y)=P(X≤x,Y≤y)=P(s(X)≤s(x),t(Y)≥t(y))

The last step is obtained by applying the functions s and t since s preserves order and t reverses it.

This can be further transformed into

F(X,Y)(x,y)=1−P(s(X)≤s(x),t(Y)≤t(y))=1−F(s(X),t(Y))(s(x),t(y))

Since our random variables are continuous, we assume that the difference between t(Y)≤t(y) and t(Y)<t(y)) is just a set of zero measure.

Now, to transform this into a statement about copulas, note that

C(X,Y)(a,b)=F(X,Y)(F−1X(a), F−1Y(b))

Thus, plugging x=F−1X(a) and y=F−1Y(b) into our previous formula,

we get

F(X,Y)(F−1X(a),F−1Y(b))=1−F(s(X),t(Y))(s(F−1X(a)),t(F−1Y(b)))

The left hand side is the copula C(X,Y), the right hand side still needs some work.

Note that

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and likewise

Ft(Y)(s(F−1Y(b)))=P(t(Y)≤t(F−1Y(b)))=P(Y≥F−1Y(b))=1−FY(F−1Y(b))=1−b

Combining all results we obtain for the relationship between the copulas

C(X,Y)(a,b)=1−C(s(X),t(Y))(a,1−b).

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