Data Sufficiency

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

gmatdriller 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

S1: P(20) = K(20-a)(20-b) = 0 Also, K(1020-a)(1020-b) = 0
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(120-a)(120-b)
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(120-20)(120-1020) = 9000
So Value of K can be found.....

Suff

IMO C

What is the OA?

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.

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.

Can you please explain.....??
Also, Why equate the derivative to zero and how to get minimum value?

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 - ∞ )
If a is positive, then the curve has a minimum value and no maximum value (technically ∞ )

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^n-1

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 2-minute 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.

eaakbari 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

not quite right, simplifying we get P(x)= k*x^2-k(a+b)*x+kab

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

Really? Have you passed calculus or failed it?
P`(x)=2kx-k(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.

(I) Clearly implies that a & b are equal to 20 and 1020
Hence we have a & b values, hence Suff

Not right again. St(1) P(20) = P(1,020) = 0 supplies as with information (remember our function is parabolic from k*x^2-k(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 in-between point between symmetric coordinates of x on the cross-over 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=(1020-20)/2=500. By plugging this into our previously found (derived) extremum point when x=(a+b)/2 we obtain a+b=1000.
Now to finalize our answer we would need individual values for a, b and k. Because to find P(500)=k*500^2-k(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*20^2-k(a+b)*20+kab=0 and P(1020)=k*1020^2-k(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)=1020^2-20^2=(1020-20)*(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).

gmatdriller 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

yes, I agree Ron. Intuitively, I have subtracted (1020-20)/2 instead of putting + in the expression. The answer is indeed C
Combined statements resolved below
st(1) P(520)=k(520-20)(520-1020)=k*500*(-500)=-250,000k
st(120)=k(120-20)(120-1020)=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