The profit function when Company C produces and sells x units of Product R is P
(x) = k(x  a)(x  b), where k, a, and b are constants and k < 0.
What is the maximum value for P(x)?
(1)P(20) = P(1,020) = 0
(2)P(120) = 9,000
The profit function when Company C
This topic has expert replies

 Master  Next Rank: 500 Posts
 Posts: 418
 Joined: 04 Jul 2010
 Thanked: 6 times
 Followed by:2 members
 nisagl750
 Master  Next Rank: 500 Posts
 Posts: 145
 Joined: 31 Jan 2012
 Location: New Delhi
 Thanked: 16 times
 Followed by:2 members
 GMAT Score:760
S1: P(20) = K(20a)(20b) = 0 Also, K(1020a)(1020b) = 0gmatdriller wrote:The profit function when Company C produces and sells x units of Product R is P
(x) = k(x  a)(x  b), where k, a, and b are constants and k < 0.
What is the maximum value for P(x)?
(1)P(20) = P(1,020) = 0
(2)P(120) = 9,000
i.e. if a=20, b=1020 Or a=1010, b=20 (Since K<0)
But no info about K....
Insuff
S2: P(120) = K(120a)(120b)
Don't know the values of a & b
Insuff
Together.
consider a=20, b=1020 (or a=1020, b=20, both are same)
P(120) = K(12020)(1201020) = 9000
So Value of K can be found.....
Suff
IMO C
What is the OA?
 eaakbari
 Master  Next Rank: 500 Posts
 Posts: 435
 Joined: 15 Mar 2010
 Thanked: 32 times
 Followed by:1 members
IMO A
I tend to use calculus to solve this type (a very basic concept of calculus though)
Info 
(x) = k(x  a)(x  b)
Simplifying
(x) = kx^2 kx(a+b) +abx
Taking derivate d(x)
d(x) = 2kx  k(b+a)
equating to zero for maximum and dividing equation by k (constant)
0 = 2x  (b+a)
x = (b+a)/2 is when x is maximum
(I) Clearly implies that a & b are equal to 20 and 1020
Hence we have a & b values, hence Suff
(II) Does not give any value for a or b. Hence Insuff
IMO A
P.S. The calculus concepts for Max & Min problems are very basic and save oodles of time. I could explain in more detail if required.
I tend to use calculus to solve this type (a very basic concept of calculus though)
Info 
(x) = k(x  a)(x  b)
Simplifying
(x) = kx^2 kx(a+b) +abx
Taking derivate d(x)
d(x) = 2kx  k(b+a)
equating to zero for maximum and dividing equation by k (constant)
0 = 2x  (b+a)
x = (b+a)/2 is when x is maximum
(I) Clearly implies that a & b are equal to 20 and 1020
Hence we have a & b values, hence Suff
(II) Does not give any value for a or b. Hence Insuff
IMO A
P.S. The calculus concepts for Max & Min problems are very basic and save oodles of time. I could explain in more detail if required.
Whether you think you can or can't, you're right.
 Henry Ford
 Henry Ford
 nisagl750
 Master  Next Rank: 500 Posts
 Posts: 145
 Joined: 31 Jan 2012
 Location: New Delhi
 Thanked: 16 times
 Followed by:2 members
 GMAT Score:760
Can you please explain.....??eaakbari wrote: P.S. The calculus concepts for Max & Min problems are very basic and save oodles of time. I could explain in more detail if required.
Also, Why equate the derivative to zero and how to get minimum value?
 eaakbari
 Master  Next Rank: 500 Posts
 Posts: 435
 Joined: 15 Mar 2010
 Thanked: 32 times
 Followed by:1 members
Calculus has got a fearsome reputation but in actuality is quite easy and has great applications.
Below I will explain enough Calculus needed to solve Maximum / Minimum problems, nothing less and nothing more.
Calculus can be expressed as the mathematical tool that studies change.
Most problem's we will encounter would be quadratic or linear equations and the question requiring us to find a max/min of the equation.
(1) ax^2 + bx + c = 0
Equation (1) is a parabolic shaped curve which has one peak value (may be positive or negative)
If a is negative, then the curve has a maximum value and no minimum value (technically  âˆž )
If a is positive, then the curve has a minimum value and no maximum value (technically âˆž )
At this point its useful to visualize on the graph and hence find the attached image
Now lets take the example of the equation which google bestowed upon me as per the image.
(2) y = x^2  4x  5
Since the coefficient of x^2 is positive (1), we know this curve has a unique minimum value.
Now lets get to the calculus of things. Remember the three easy formulae as below and your life is sorted.
For these questions, only the concept of differential calculus is required.
Derivative of x^n is :
(a) d (x^n) = n.x^n1
where d denotes differential or derivative
Derivative of constant is 0:
(b) d (c) = 0
where d denotes differential or derivative
Derivative of function with a constant is:
d (a.f(x)) = a.f'(x)
where a is the constant and f' is the derivative.
And voila! That's all you need to know about Calculus (for now at least)
For any maximum or minimum value, equate the derivative of the equation to zero and you will get the max or min.
Taking equation (2)
(2) y = x^2  4x  5
Using the formula (a) & (b), we get
y' = 2*x  4  0
y' = 2*x  4
Equation y' to 0
0 = 2*x  4
x = 2
Hence y is minimum when x is 2.
Using this you can solve a 2minute question in 15 seconds
Unfortunately I have just started studying for the GMAT and don't have any example question as of now to to explain this in a real GMAT problem. Please post one, so that we can solve it.
Below I will explain enough Calculus needed to solve Maximum / Minimum problems, nothing less and nothing more.
Calculus can be expressed as the mathematical tool that studies change.
Most problem's we will encounter would be quadratic or linear equations and the question requiring us to find a max/min of the equation.
(1) ax^2 + bx + c = 0
Equation (1) is a parabolic shaped curve which has one peak value (may be positive or negative)
If a is negative, then the curve has a maximum value and no minimum value (technically  âˆž )
If a is positive, then the curve has a minimum value and no maximum value (technically âˆž )
At this point its useful to visualize on the graph and hence find the attached image
Now lets take the example of the equation which google bestowed upon me as per the image.
(2) y = x^2  4x  5
Since the coefficient of x^2 is positive (1), we know this curve has a unique minimum value.
Now lets get to the calculus of things. Remember the three easy formulae as below and your life is sorted.
For these questions, only the concept of differential calculus is required.
Derivative of x^n is :
(a) d (x^n) = n.x^n1
where d denotes differential or derivative
Derivative of constant is 0:
(b) d (c) = 0
where d denotes differential or derivative
Derivative of function with a constant is:
d (a.f(x)) = a.f'(x)
where a is the constant and f' is the derivative.
And voila! That's all you need to know about Calculus (for now at least)
For any maximum or minimum value, equate the derivative of the equation to zero and you will get the max or min.
Taking equation (2)
(2) y = x^2  4x  5
Using the formula (a) & (b), we get
y' = 2*x  4  0
y' = 2*x  4
Equation y' to 0
0 = 2*x  4
x = 2
Hence y is minimum when x is 2.
Using this you can solve a 2minute question in 15 seconds
Unfortunately I have just started studying for the GMAT and don't have any example question as of now to to explain this in a real GMAT problem. Please post one, so that we can solve it.
Whether you think you can or can't, you're right.
 Henry Ford
 Henry Ford
not quite right, simplifying we get P(x)= k*x^2k(a+b)*x+kabeaakbari wrote:IMO A
I tend to use calculus to solve this type (a very basic concept of calculus though)
Info 
(x) = k(x  a)(x  b)
Simplifying
(x) = kx^2 kx(a+b) +abx
Really? Have you passed calculus or failed it?d(x) = 2kx  k(b+a)
equating to zero for maximum and dividing equation by k (constant)
0 = 2x  (b+a)
x = (b+a)/2 is when x is maximum
P`(x)=2kxk(a+b)=0, 2kx=k(a+b), 2x=a+b, x=(a+b)/2 <== This is the extremum point (where function P may get minimum or maximum value, we don't know yet).
Take the second derivative d^2P/dx^2=P``(x) or 2k. Now depending on the sign of k we get close to our answer. If k<0, we will have maximum value in the point x=(a+b)/2 and by substituting this into P(x) we get P(x) maximum.
The condition in the question says that k<0, hence P is maximized in the point x=(a+b)/2.
Not right again. St(1) P(20) = P(1,020) = 0 supplies as with information (remember our function is parabolic from k*x^2k(a+b)*x+kab, where a,b,k are constants and k cancelled out leaves us with x^2(a+b)x+ab, similar to GMAT Foiler) that there's some inbetween point between symmetric coordinates of x on the crossover of y=P(x)=0. That is the half of interval [20;1020] will be our maximum point. Hence our function will be maximzed at x=(102020)/2=500. By plugging this into our previously found (derived) extremum point when x=(a+b)/2 we obtain a+b=1000.(I) Clearly implies that a & b are equal to 20 and 1020
Hence we have a & b values, hence Suff
Now to finalize our answer we would need individual values for a, b and k. Because to find P(500)=k*500^2k(1000)*500+kab we are missing exactly a,b,k.
Since in St(1) we are given P(20)=0 and P(1020)=0 we may attempt to set several equations to find the three unknown variables (a,b,k).
P(20)=k*20^2k(a+b)*20+kab=0 and P(1020)=k*1020^2k(a+b)*1020+kab=0. We know that k is not 0 and we are left with 20^2(a+b)*20+ab=1020^2(a+b)*1020+ab. Cancelling ab on both sides we get, 20^2(a+b)*20=1020^2(a+b)*1020 or 1040*(a+b)=1020^220^2=(102020)*(1020+20)=1000*1040 and a+b=1000
I have gone through so much tedious explanation with calculations to show that St(1) is not sufficient, i.e. St(1) In Not Right.
St(2) P(120) = 9,000 Cannot be right as well, as we are put to find the values of a,b,k again. So St(2) alone is not sufficient.
Combining both statements(1&2), we still have only two equations on the side of St(1) which are equivalent (we have seen that) and one equation on the side of St(2). So three variables and two equations. Unless we introduce a new parameter, this may not solvable. The parametric equation will not indicate one unique answer, rather the set of values underpinned by parameter (GMAT wants one, unique answer)
Answer is E (the question is not solvable with any statement taken alone, neither combined).
Success doesn't come overnight!
GMAT/MBA Expert
 lunarpower
 GMAT Instructor
 Posts: 3380
 Joined: 03 Mar 2008
 Thanked: 2256 times
 Followed by:1524 members
 GMAT Score:800
i got a private message about this thread.
this problem is very much unlike anything i've ever seen in an official problem, so the best advice is to ignore it.
for what it's worth, a parabola is uniquely determined by any 3 points, so the answer to this problem is (c).
this problem is very much unlike anything i've ever seen in an official problem, so the best advice is to ignore it.
for what it's worth, a parabola is uniquely determined by any 3 points, so the answer to this problem is (c).
Ron has been teaching various standardized tests for 20 years.

Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions Ã Ron en franÃ§ais
Voit esittÃ¤Ã¤ kysymyksiÃ¤ Ron:lle myÃ¶s suomeksi

Quand on se sent bien dans un vÃªtement, tout peut arriver. Un bon vÃªtement, c'est un passeport pour le bonheur.
Yves SaintLaurent

Learn more about ron

Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions Ã Ron en franÃ§ais
Voit esittÃ¤Ã¤ kysymyksiÃ¤ Ron:lle myÃ¶s suomeksi

Quand on se sent bien dans un vÃªtement, tout peut arriver. Un bon vÃªtement, c'est un passeport pour le bonheur.
Yves SaintLaurent

Learn more about ron
GMAT/MBA Expert
 lunarpower
 GMAT Instructor
 Posts: 3380
 Joined: 03 Mar 2008
 Thanked: 2256 times
 Followed by:1524 members
 GMAT Score:800
also  as far as statement 1 is concerned, i don't understand the point of all the calculus.
those two given values can just be substituted directly, to give 0 = k(20  a)(20  b) and 0 = k(1020  a)(1020  b). if those are true, then it follows at once that "a" and "b" are, in some order, 20 and 1020.

pemdas 
the halfway point between 20 and 1020 is 520, not 500.
you don't subtract to find a midpoint. you just take the average (i.e., add and divide by 2).
if it's not clear why, then look at simpler numbers.
e.g., what's the halfway point between 8 and 10?
by inspection the answer should clearly be 9. if you take the average, you get (8 + 10)/2 = 9, as expected.
on the other hand, this subtraction thing would give (10  8)/2 = 1, which is definitely not the midpoint between 8 and 10.
those two given values can just be substituted directly, to give 0 = k(20  a)(20  b) and 0 = k(1020  a)(1020  b). if those are true, then it follows at once that "a" and "b" are, in some order, 20 and 1020.

pemdas 
the halfway point between 20 and 1020 is 520, not 500.
you don't subtract to find a midpoint. you just take the average (i.e., add and divide by 2).
if it's not clear why, then look at simpler numbers.
e.g., what's the halfway point between 8 and 10?
by inspection the answer should clearly be 9. if you take the average, you get (8 + 10)/2 = 9, as expected.
on the other hand, this subtraction thing would give (10  8)/2 = 1, which is definitely not the midpoint between 8 and 10.
Ron has been teaching various standardized tests for 20 years.

Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions Ã Ron en franÃ§ais
Voit esittÃ¤Ã¤ kysymyksiÃ¤ Ron:lle myÃ¶s suomeksi

Quand on se sent bien dans un vÃªtement, tout peut arriver. Un bon vÃªtement, c'est un passeport pour le bonheur.
Yves SaintLaurent

Learn more about ron

Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions Ã Ron en franÃ§ais
Voit esittÃ¤Ã¤ kysymyksiÃ¤ Ron:lle myÃ¶s suomeksi

Quand on se sent bien dans un vÃªtement, tout peut arriver. Un bon vÃªtement, c'est un passeport pour le bonheur.
Yves SaintLaurent

Learn more about ron
GMAT/MBA Expert
 ceilidh.erickson
 GMAT Instructor
 Posts: 2094
 Joined: 04 Dec 2012
 Thanked: 1443 times
 Followed by:245 members
Just to add to what Ron said... you never need to use calculus on the GMAT! There may be a small handful of problems that could have calculuslike applications, but there's always another (and almost certainly easier) way to solve.
So this is a publicservice plea to posters out there: please do not post questions on this forum that are unlike real GMAT questions! It can certainly be fun to challenge yourself with other hard questions, but this isn't the place for it. For one, it can scare or mislead other students who aren't proficient in calculus, and who don't need to be for the GMAT! For another, it doesn't help you to build skills that will be applicable to the GMAT.
When you're posting, LIST YOUR SOURCE, so the rest of us can gage how reliable it is as a GMAT source. Please don't intentionally post anything that wasn't meant to be a GMAT question.
So this is a publicservice plea to posters out there: please do not post questions on this forum that are unlike real GMAT questions! It can certainly be fun to challenge yourself with other hard questions, but this isn't the place for it. For one, it can scare or mislead other students who aren't proficient in calculus, and who don't need to be for the GMAT! For another, it doesn't help you to build skills that will be applicable to the GMAT.
When you're posting, LIST YOUR SOURCE, so the rest of us can gage how reliable it is as a GMAT source. Please don't intentionally post anything that wasn't meant to be a GMAT question.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
yes, I agree Ron. Intuitively, I have subtracted (102020)/2 instead of putting + in the expression. The answer is indeed Cgmatdriller wrote:The profit function when Company C produces and sells x units of Product R is P
(x) = k(x  a)(x  b), where k, a, and b are constants and k < 0.
What is the maximum value for P(x)?
(1)P(20) = P(1,020) = 0
(2)P(120) = 9,000
Combined statements resolved below
st(1) P(520)=k(52020)(5201020)=k*500*(500)=250,000k
st(120)=k(12020)(1201020)=9,000 and k=9,000/9,000=1
P(520)=1*(250,000)=250,000 Hence profit is maximised as the value of 250,000
Success doesn't come overnight!
GMAT/MBA Expert
 lunarpower
 GMAT Instructor
 Posts: 3380
 Joined: 03 Mar 2008
 Thanked: 2256 times
 Followed by:1524 members
 GMAT Score:800
ok.
well, just remember the 2 most important things here:
1/
you will never see a problem like this on the gmat.
except for the fact that it's written in DS format, this problem looks like something taken directly from a firstyear calculus textbook. the GMAT problems come in lots of different flavors, but they never resemble routine "textbook" problems.
2/
if you have any quadratic function and you have two inputs that give the same value, then the max/min of the function is always halfway between those points.
there's no need to throw tons of work at this; if you see f(a) = f(b) and f is a quadratic function, then the max/min value will always occur at (a + b)/2.
well, just remember the 2 most important things here:
1/
you will never see a problem like this on the gmat.
except for the fact that it's written in DS format, this problem looks like something taken directly from a firstyear calculus textbook. the GMAT problems come in lots of different flavors, but they never resemble routine "textbook" problems.
2/
if you have any quadratic function and you have two inputs that give the same value, then the max/min of the function is always halfway between those points.
there's no need to throw tons of work at this; if you see f(a) = f(b) and f is a quadratic function, then the max/min value will always occur at (a + b)/2.
Ron has been teaching various standardized tests for 20 years.

Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions Ã Ron en franÃ§ais
Voit esittÃ¤Ã¤ kysymyksiÃ¤ Ron:lle myÃ¶s suomeksi

Quand on se sent bien dans un vÃªtement, tout peut arriver. Un bon vÃªtement, c'est un passeport pour le bonheur.
Yves SaintLaurent

Learn more about ron

Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions Ã Ron en franÃ§ais
Voit esittÃ¤Ã¤ kysymyksiÃ¤ Ron:lle myÃ¶s suomeksi

Quand on se sent bien dans un vÃªtement, tout peut arriver. Un bon vÃªtement, c'est un passeport pour le bonheur.
Yves SaintLaurent

Learn more about ron