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Finite Math

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User Generated
Subject
Mathematics
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Homework
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The total payment is
12 5
0.031
1 199,000 1
1
232,317.50
2
nt
r
AP
n
= + = +
=
. It follows that the monthly
payment must be
232,317.50
$3871.96
5 6012
A
==
.

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250 10500P x y=+
.
The second and third constraints imply the first one,
2
4600
2000
( 10
15
2
)
x
x
yx
y
y
+
+
+
.
The feasible region marked by yellow.

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nt 125  r  0.031  The total payment is A = P 1 +  = 199, 000 1 +  12   n  A 232,317.50 = = $3871.96 . payment must be 12  5 60 = 232,317.50 . It follows that the monthly P = 250 x + 10500 y . The second and third constraints imply the first one,  x + 15 y  4600  x + 2 y  2000 .  2  y  ( x + 10) The feasible region marked by yellow.  250  1 The gradient of the profit function is P =   = 250   . We graph it up to scale by red. 10500   42  4600 2 = 306 . 15 3 The IP-solution can be derived by y0 = 306 and x0 =  306  − 10 = 17 . The solution of LP-problem is x = 0 and y = The maximal profit equals 250 17 + 10500  306 = 3, 217, 250 (a) ( x + 590) + ( y + 590) − 590 = 740 + 840 − 590 = 990 ; (b) z = 1200 − 990 = 210 ; (c) y = 840 − 590 = 250 . (a)  A  B = 26 + 20 + 14 = 60 ; (b)  A = 40 + 20 = 60 ;  A  B  = 60 = 0.6 ; U  60 + 40  A  B  = 1 − 14 = 0.86 ; (d) P ( ( A  B )  ) = 1 − P ( A  B ) = 1 − 100 U  (e) ( A  B) +  A  B − 46 = U  −  A  B +  A  B − 46 = U  −14 + 60 − 46 = U  (c) P ( A  B ) = Name: Finite Math Description: The second and third constraints imply the first one, The IP-solution can be derived by y0 = 306 and x0 = ?? 306 ?? ? 10 = 17 . (e) ?( A ... ...
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