The daily profit, P, for selling x units of a certain item at

Question

The daily profit, P, for selling x units of a certain item at a sporting goods store can be modeled by the function  P(x) = -a(x -\frac{b}{2a})^2 + \frac{b^2}{4a}+c , where a and b are positive constants and c is a nonnegative constant. What is the maximum daily profit for selling this item?

(1)  b^2 + 4ac = \frac{52ac}{3}

(2) c = 360

KarishmaB · Accepted Answer

Note the 3 terms of the profit expression carefully:

P(x) = -a(x -\frac{b}{2a})^2 + \frac{b^2}{4a}+c

'a' and 'b' are both positive so 'a' multiplied by a square will be positive (if square is not 0) which means the first term   -a(x -\frac{b}{2a})^2  is negative or 0. For maximum profit, this term should be 0. The other two terms are certainly positive or 0.

Maximum P = \frac{b^2}{4a}+c = \frac{b^2+4ac}{4a}

(1)  b^2 + 4ac = \frac{52ac}{3}

Maximum P = \frac{52ac}{3*4a} = \frac{52c}{3*4} 
We don't know the value of c so not sufficient.

(2) c = 360

We don't know the value of a and b so not sufficient.

Using both,  Maximum P = \frac{52c}{3*4} = \frac{52*360}{3*4}

We can find this value and hence it is sufficient.

Answer (C)

Kinshook · Answer

Given: The daily profit, P, for selling x units of a certain item at a sporting goods store can be modeled by the function P(x) = -a(x -\frac{b}{2a})^2 + \frac{b^2}{4a}+c , where a and b are positive constants and c is a nonnegative constant. Asked: What is the maximum daily profit for selling this item?
Since a is positive, the maximum value of -a(x -\frac{b}{2a})^2 = 0 since -a is negative and the value of expression is 0 at x =\frac{b}{2a}
Maximum profit P(x=b/2a) = \frac{b^2}{4a}+c (1) b^2 + 4ac = \frac{52ac}{3} Maximum daily profit = \frac{b^2}{4a}+c = \frac{bˆ2 + 4ac }{ 4a} = \frac{52ac}{3*4a} = \frac{13c}{3} Since the value of c is unknown NOT SUFFICIENT (2) c = 360 Maximum daily profit = \frac{b^2}{4a}+c = \frac{b^2}{4a} + 360 Since value of \frac{b^2}{4a} is unknown NOT SUFFICIENT (1) + (2) (1) b^2 + 4ac = \frac{52ac}{3} Maximum daily profit = \frac{b^2}{4a}+c = \frac{bˆ2 + 4ac }{ 4a} = \frac{52ac}{3*4a} = \frac{13c}{3} (2) c = 360 Maximum daily profit = \frac{13c}{3} = \frac{13*360}{3} = 13*120 = 1560 SUFFICIENT IMO C

Bunuel · Answer

Is this question from the Officil Guide or GMAT Prep Focus exams? Thank you.

HarshaBujji · Answer

-(x-a)^2 + K   will be maximum at x = a, 
Above equation

P(x)=−a(x−b/2a)^2+b^2/4a+c , This equation can be represented in the above format as the  b^2/4a+c  will always be a constant.

As -a(k)^2 will always be a negtive number, P(x) will be maximum at x=b/2a, Making the negitive componant to be 0.

Hence max value of p(x) =  b^2/4a+c

To get the value of this we need to use both Stmt 1 & Stmt 2.

Hence IMO C.

MT1302 · Answer

chetan2u   KarishmaB   Bunuel   MartyMurray

Kindly share your method of solving this question.

ilikeshoppingalot · Answer

Bunuel  this is from the official mock test #5

Bunuel · Answer

Changed the source to GMAT Prep (Focus). Thank you!

Bing_Yan · Answer

I do think there is mistake in the final answer C.

Let's look at the question and it said " for selling X units of a cerain item at a sporting goods store", which clearly indicate X is a positive integer because usually we cannot say that we sell 1.5 units of sport item. 
At the same time, we know when x=b/2a, P would achieve the maximum profit. However, we cannot ensure that when x=b/2a, b/2a is a integer because question only mentioned a and b are postive constants. Therefore, mathematically speaking, when x=b/2a, P achieves the maximum profit but in this case x probably cannot exactly equal to b/2a. x only could equal to the integer near to b/2a.

Therefore, actually we don't know what's the nearest integer number to b/2a, then we cannot know what's the maximum daily profit.

So i think E should be right answer.

Bing_Yan · Answer

I do think there is mistake in the final answer C.

Let's look at the question and it said " for selling X units of a cerain item at a sporting goods store", which clearly indicate X is a positive integer because usually we cannot say that we sell 1.5 units of sport item. 
At the same time, we know when x=b/2a, P would achieve the maximum profit. However, we cannot ensure that when x=b/2a, b/2a is a integer because question only mentioned a and b are postive constants. Therefore, mathematically speaking, when x=b/2a, P achieves the maximum profit but in this case x probably cannot exactly equal to b/2a. x only could equal to the integer near to b/2a.

Therefore, actually we don't know what's the nearest integer number to b/2a, then we cannot know what's the maximum daily profit.

So i think E should be right answer.

bumpbot · Answer

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The daily profit, P, for selling x units of a certain item at

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