Find three positive numbers x, y, and z whose sum is 465 such that x^a*y^b*z^c is a maximum.
Yes, I know that a, b, and c aren't given. That's the whole point of the problem.
Yes, I know that a, b, and c aren't given. That's the whole point of the problem.
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If any one of a, b, or c is negative, then there is no maximum. To see this, suppose that a < 0. Then we can take x as close to zero as we want and drive the product x^a y^b z^c as high as we want.
If each of a = 0, b = 0, and c = 0, then the product is 1 independent of the values of x, y, and z.
So, we can consider a, b, c ≥ 0 with at least one of a, b, and c nonzero. Since x, y, and z are positive, we have no fear of division by zero. Using Lagrange multipliers we need to solve the system
a x^(a-1)y^bz^c = λ
b x^ay^(b - 1)z^c = λ
c x^ay^bz^(c - 1) = λ
x + y + z = 465.
The first three equations give
a/x = b/y = c/z = λ.
If any one of a, b, or c is zero, they must all be zero. That case has already been considered and isn't particularly interesting.
If we suppose that each of a, b, and c is positive, then the above can be rewritten as
x/a = y/b = z/c ==> x = (a/c)z and y = (b/c)z.
Using the constraint function, this gives
x = 465a/(a + b + c), y = 465b/(a + b + c), and z = 465c/(a + b + c).
If each of a = 0, b = 0, and c = 0, then the product is 1 independent of the values of x, y, and z.
So, we can consider a, b, c ≥ 0 with at least one of a, b, and c nonzero. Since x, y, and z are positive, we have no fear of division by zero. Using Lagrange multipliers we need to solve the system
a x^(a-1)y^bz^c = λ
b x^ay^(b - 1)z^c = λ
c x^ay^bz^(c - 1) = λ
x + y + z = 465.
The first three equations give
a/x = b/y = c/z = λ.
If any one of a, b, or c is zero, they must all be zero. That case has already been considered and isn't particularly interesting.
If we suppose that each of a, b, and c is positive, then the above can be rewritten as
x/a = y/b = z/c ==> x = (a/c)z and y = (b/c)z.
Using the constraint function, this gives
x = 465a/(a + b + c), y = 465b/(a + b + c), and z = 465c/(a + b + c).
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Let P=x^a*y^b*z^c
Ln(P)=aln(x)+bln(y)+cln(z)
Maximizing Ln(P) is the same as maximizing P, since ln(P) is monotonically increasing
Form the Lagrangian
L=Ln(P)-k(x+y+z-465)
differentiate by x,y,z separately and equate to zero
dL/dx=a/x-k=0, so x=a/k
similarly, y=b/k and z=c/k
But x+y+z=465, so (a+b+c)/k=465 and k=(a+b+c)/465
So the answer is
x=465a/S, y=465b/S and z=465c/S where S=a+b+c
Ln(P)=aln(x)+bln(y)+cln(z)
Maximizing Ln(P) is the same as maximizing P, since ln(P) is monotonically increasing
Form the Lagrangian
L=Ln(P)-k(x+y+z-465)
differentiate by x,y,z separately and equate to zero
dL/dx=a/x-k=0, so x=a/k
similarly, y=b/k and z=c/k
But x+y+z=465, so (a+b+c)/k=465 and k=(a+b+c)/465
So the answer is
x=465a/S, y=465b/S and z=465c/S where S=a+b+c