The Marquis de America asked:
Optimization function to be written as function to be optimized what is the feasible domain of which should be written as function to be written as function to be optimized what.
Optimization function to be written as function to be optimized what is the feasible domain of which should be written as function to be written as function to be optimized what.
Let a,b be two given numbers.
ab=192.
b=192/a.
Let f ;N–>N,where N denote set of the natural numbers.
f(a)=a+b=a+192/a=y
a^2–ya+192=0.
We should have y^2-4×192>=0.[and it should be a perfect square].
y=16(3)^1/2.
Since y=a+b is a perfect square.
Y=32 is the minimum natural number.[sum]
a+b=32,
The numbers are 8&24.
The domain is N.
July 3rd, 2009 at 3:54 am
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The two numbers are 12 and 16
The two equations for part ‘a’ are:
(1) xy = (3•2^n)(2^(6-n))
(2) S(n) = (3•2^n) + 2^(6-n)
For part ‘b’:
The Domain of S(n) = {n| n is an integer and 0 ? n ? 6}
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You’re right. This is a tricky problem.
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Find two positive numbers with a product of 192 such that their sum is a minimum.
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STEP 1
If you prime factor 192, you get 192 = 3•2^6
Since we are looking for two numbers and there is only one three, one of the numbers will contain the factor 3 and its share of the 2’s. The other number will then be a power of 2.
Let the two numbers be x and y. Then,
xy = 192
xy = 3•2^6
xy = 3(2^n)[2^(6-n)]
xy = [3•2^n][2^(6-n)]…This is an answer to Part a
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STEP 2
Let x = 3•2^n
Let y = 2^(6-n)
Let S(n) = x + y
……S(n) = (3•2^n) + 2^(6-n)…This is the optimization function for Part a
Taking the derivative gets us
S’ = 3[(2^n)(dn/dn)(ln2)] + [2^(6-n)](-1)(ln2)…this simplifies to
S’ = [3(2^n) - 2^(6-n)](ln2)
The minimum will occur when S’ = 0. And, S’ = 0 when
3(2^n) - 2^(6-n) = 0
…………….3•2^n = 2^(6-n)…now, take the log of both sides and simplify
………..ln(3•2^n) = ln(2^(6-n))
…….ln3 + n(ln2) = (6-n)(ln2)
…………………ln3 = (6-n)(ln2) - n(ln2)
…………………ln3 = (6-n-n)(ln2)
…………………ln3 = 2(3-n)(ln2)
……..ln3 / 2(ln2) = 3-n
…………………..n = 3 - [ln3 / 2(ln2)]
…………………..n ? 2
…………………..n = 2
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So, the two numbers are:
x = 3(2^n) = 3•2² = 12
y = 2^(6-n) = 2^(6-2) = 2^4 = 16
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Here’s an extra. All the possible pairings for x and y
X……3……6……12…..24
Y….64…..32……16……8
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X….48…..96….192
Y….. 4…….2……..1
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July 4th, 2009 at 4:15 pm
qa
xy = 192
y = 192 / x
f(x) = x + y
= x + 192/x
qb
R > 0.
12 and 16 if you wanted the domain be N.
x = y = 8√3
July 7th, 2009 at 6:14 pm