Did you make up that question, or does it come from a particular source?oldheaven wrote:If 5xy+9yz+2xz=30 , then what is the maximum of xyz ?
1)100
2)100/3
3)10/3
4)3
Maximum
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tomada
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user123321
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should be 10/3oldheaven wrote:If 5xy+9yz+2xz=30 , then what is the maximum of xyz ?
1)100
2)100/3
3)10/3
4)3
we know that AM >= GM
so if we consider the terms of the series as 5xy,9yz & 2xz and apply the above principle
(5xy+9yz+2xz)/3 >= (5xy.9yz.2xz)^(1/3)
=> (5xy+9yz+2xz)/3 >= (90.x^2.y^2.z^2)^(1/3) - (1)
but we know
5xy+9yz+2xz=30 from given question
substitute in equation (1)
30/3 > = (90.x^2.y^2.z^2)^(1/3)
10 >= (90.x^2.y^2.z^2)^(1/3)
cubing on both sides
1000 >= 90.x^2.y^2.z^2
=>xyz <= 10/3
=> max of xyz is 10/3
hope this helps
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pemdas
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it's obvious, y and z should be minimized and x maximized -> y=z=1 and 5x+9+2x=30, 7x=21, x=3
and xyz=3
4
and xyz=3
4
oldheaven wrote:If 5xy+9yz+2xz=30 , then what is the maximum of xyz ?
1)100
2)100/3
3)10/3
4)3
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LalaB
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why is that? why do u minimize exactly y and z?because of stated answer choices?pemdas wrote:it's obvious, y and z should be minimized and x maximized
imho, the q. is not a gmat q.
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pemdas
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no, because we are dealing with polynomials and I'm assessing the coefficients of the given polynomials. By minimizing the highest coefficient polynomials I am increasing the value of xyz.
LalaB wrote:why is that? why do u minimize exactly y and z?because of stated answer choices?pemdas wrote:it's obvious, y and z should be minimized and x maximized
imho, the q. is not a gmat q.
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tomada
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Pemdas, the assumptions that y=z=1 are valid only if we are told that x,y,z are integers > 0.pemdas wrote:it's obvious, y and z should be minimized and x maximized -> y=z=1 and 5x+9+2x=30, 7x=21, x=3
and xyz=3
4oldheaven wrote:If 5xy+9yz+2xz=30 , then what is the maximum of xyz ?
1)100
2)100/3
3)10/3
4)3
otherwise, they can assume much smaller values...
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user123321
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Take any set of non -ve numbers,oldheaven wrote:I don't understand what exactly it means!we know that AM >= GM
There is a proven theorem which says that always
Arthmetic Mean(AM) will be greater than or equal to Geometric Mean(GM).
src: https://en.wikipedia.org/wiki/Inequality ... tric_meansIn mathematics, the inequality of arithmetic and geometric means, or more briefly the AM-GM inequality, states that the arithmetic mean of a list of non-negative real numbers is greater than or equal to the geometric mean of the same list; and further, that the two means are equal if and only if every number in the list is the same.
you can try with few numbers.
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pemdas
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oldheaven, are you sure we should be preped for GM included as concept with this q.?
i barely remember this concept, not to speak about its application for test
tomada, i totally agree with you here
again, i would mark xyz=3 if sitting for GMAT and using only polynomials background
thanks to oldheaven and user we remembered now some math/stats concepts
i barely remember this concept, not to speak about its application for test
tomada, i totally agree with you here
see user also assumed non -ve values, the condition for GM as product of the substitutable is taken under root which may be even ...tomada wrote: Pemdas, the assumptions that y=z=1 are valid only if we are told that x,y,z are integers > 0.
otherwise, they can assume much smaller values...
again, i would mark xyz=3 if sitting for GMAT and using only polynomials background
thanks to oldheaven and user we remembered now some math/stats concepts
Success doesn't come overnight!
