What's the OA?

oldheaven wrote:x,y>0 .If xy=6 then what is the minimum of 3x+5y ?

1)6sqrt(10)
2)sqrt(30)
3)2sqrt(10)
4)3aqrt(6)

Please don't shower bats at me for solving this question, this way !

If x,y>0, xy=6 then what is the minimum of 3x+5y ?

1)6sqrt(10)
2)sqrt(30)
3)2sqrt(10)
4)3aqrt(6)

xy = 6, Implies y =6/x
The expression 3x + 5y = 3x + (30/x) = (3x^2 + 30)/x
Maximum/Minimum value of a function f(x) can be found by initially finding the value of x that satisfies the equation f '(x) = 0 (where f '(x) is the derivative of f(x)) and then substitution the value of x in the function f(x).

f(x) = (3x^2 + 30)/x
f'(x) = (3x^2 - 30)/x^2 = 0 ; x =√10 and the derivative of f'(x)>0(which tells us that f(x) at x =√10 is minimum)
The expression 3x + 5y = 3x + (30/x) = 3√10 +(30/√10) = 6√10 A

Derivative formulae -> https://math.about.com/library/weekly/aa021003a.htm- I used rule #9

I love it! Well done! Do we need to know calculus for the GMAT?

neelgandham wrote:Please don't shower bats at me for solving this question, this way !

If x,y>0, xy=6 then what is the minimum of 3x+5y ?

1)6sqrt(10)
2)sqrt(30)
3)2sqrt(10)
4)3aqrt(6)

xy = 6, Implies y =6/x
The expression 3x + 5y = 3x + (30/x) = (3x^2 + 30)/x
Maximum/Minimum value of a function f(x) can be found by initially finding the value of x that satisfies the equation f '(x) = 0 (where f '(x) is the derivative of f(x)) and then substitution the value of x in the function f(x).

f(x) = (3x^2 + 30)/x
f'(x) = (3x^2 - 30)/x^2 = 0 ; x =√10 and the derivative of f'(x)>0(which tells us that f(x) at x =√10 is minimum)
The expression 3x + 5y = 3x + (30/x) = 3√10 +(30/√10) = 6√10 A

Derivative formulae -> https://math.about.com/library/weekly/aa021003a.htm- I used rule #9

Thanks Tomada ! and I am sure that you need not have knowledge of calculus for the GMAT(Knowing some formulae doesn't hurt, and sometimes helps as in this case). I am sure we can solve this in a better(read using algebra) way.So, let us wait for the experts to reply!

neelgandham wrote:Thanks Tomada ! and I am sure that you need not have knowledge of calculus for the GMAT(Knowing some formulae doesn't hurt, and sometimes helps as in this case). I am sure we can solve this in a better(read using algebra) way.So, let us wait for the experts to reply!

There is an algebraic solution, but this still seems to be a very unGMATlike question in that it only has four answer choices and is extremely difficult. I think that people who are only interested in the GMAT would be wise to skip this question.

I think the more important matter, though, is the origin and meaning of the phrase "please don't shower bats at me"

oldheaven wrote:x,y>0 .If xy=6 then what is the minimum of 3x+5y ?

1)6sqrt(10)
2)sqrt(30)
3)2sqrt(10)
4)3aqrt(6)

Okay I just read some of your other posts explaining the source of these questions. You say it's some kind of GMAT-like test in Iran? If you post a non-standard GMAT question, I think it's important to note the source so that other people don't get worried if it seems way too hard for them.

That said, one way to solve this algebraically is as follows:

xy=6 and we're trying to minimize the value of 3x+5y. Let 3x+5y=b. So if we consider all the possible values of x and y that satisfy xy=6, we can plug these into 3x+5y=b, which would give us a value b. Our job is to find the lowest possible value of b. xy=6---->y=6/x. Plug this in to 3x+5y=b to get 3x+5(6/x)=b:

3x+30/x=b

Multiply both sides by x:

3x^2+30=bx

3x^2-bx+30=0

For this to have a solution, the discriminant must be non-negative, so

b^2-4ac>=0

b^2-4(3)(30)>=0

b^2-360>=0

b^2>=360

We're only considering x,y>0, so

b>=√360

b>=6√10

yes, you are right and I should have mentioned the source.your approach to solve the question is fantastic.I really really enjoyed it and I appreciate you.

neelgandham wrote:Please don't shower bats at me for solving this question, this way !

If x,y>0, xy=6 then what is the minimum of 3x+5y ?

1)6sqrt(10)
2)sqrt(30)
3)2sqrt(10)
4)3aqrt(6)

xy = 6, Implies y =6/x
The expression 3x + 5y = 3x + (30/x) = (3x^2 + 30)/x
Maximum/Minimum value of a function f(x) can be found by initially finding the value of x that satisfies the equation f '(x) = 0 (where f '(x) is the derivative of f(x)) and then substitution the value of x in the function f(x).

f(x) = (3x^2 + 30)/x
f'(x) = (3x^2 - 30)/x^2 = 0 ; x =√10 and the derivative of f'(x)>0(which tells us that f(x) at x =√10 is minimum)
The expression 3x + 5y = 3x + (30/x) = 3√10 +(30/√10) = 6√10 A

Derivative formulae -> https://math.about.com/library/weekly/aa021003a.htm- I used rule #9

and to be more precise with calculus ...
f(x) = (3x^2 + 30)/x = 3x + 30*x^-1 and df/dx = f`(x) = 3 - 30/x^2
Equating this to 0 (zero) implies either maximum or minimum => 3 - 30/x^2 = 0; 3 = 30/x^2, x^2= 10 and x=sqrt(10). To understand this is minimum of f(x) we take the second derivative, d^2f/dx^2=f``(x)=(3-30/x^2)'=(-30*x^-2)'=60/x^3. Since x>0 60/x^3>0 and we have minimum of f(x) in the point x=sqrt(10). No we supply this into original function f(x) = (3x^2 + 30)/x and find f(x) = (30+30)/sqrt(10)=60/sqrt(10)=6sqrt(10)

Hi Brent, guess Anil and I are out of GMATing too

Brent@GMATPrepNow wrote:Since this thread is still alive and kicking, I want to make it clear that the question is out of scope.

Cheers,
Brent

Thanks Brent your post gave some relief to me.

Vivek.