Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 15 Jan 2009
Posts: 43
Schools: Marshall '11

In a certain company, the formula for maximizing profits is
[#permalink]
Show Tags
Updated on: 26 Jan 2014, 03:21
Question Stats:
59% (01:35) correct 41% (02:01) wrong based on 485 sessions
HideShow timer Statistics
In a certain company, the formula for maximizing profits is P = 25x^2 + 7500x,, where P is profit and x is the number of machines the company operates in its factory. What value for x will maximize P? A) 10 B) 50 C) 150 D) 200 E) 300 This one is from IntegratedLearning. Here's what I did: P=25x2+7500X => P=50+7500X, to max. P, we need to max X. So I picked E, why isn't that correct?
The OA states: To find a maximum or minimum value of an equation with an exponent in it, you take the derivative of the equation, set it to zero, and solve. I don't really get what that means. So whoever solves it, could you plz post explanation of what the above sentence mean as well?
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by wcgc on 09 Feb 2009, 02:12.
Last edited by Bunuel on 26 Jan 2014, 03:21, edited 1 time in total.
Edited the question.




Math Expert
Joined: 02 Sep 2009
Posts: 62499

Re: Finding Max or Min value of equation
[#permalink]
Show Tags
14 Aug 2010, 06:32
appy001 wrote: Thanks for this post to have a recall of calculus..."What value for x will maximize P" Need more clarification on minimum value....
what if require the value of x when profit is lowest???
My understanding is that it comes from the formula for minimize profit. This will be given in question stem... such as this is present here... "the formula for maximizing profits is P = 25x2 + 7500x"....
Please put the correct understanding here. Couple of things: Quadratic expression \(ax^2+bx+c\) reaches its extreme values when \(x=\frac{b}{2a}\). When \(a>0\) extreme value is minimum value of \(ax^2+bx+c\) (maximum value is not limited) and when \(a<0\) extreme value is maximum value of \(ax^2+bx+c\) (minimum value is not limited). You can look at this geometrically: \(y=ax^2+bx+c\) when graphed on XY plane gives parabola. When \(a>0\), the parabola opens upward and minimum value of \(ax^2+bx+c\) is ycoordinate of vertex, when \(a<0\), the parabola opens downward and maximum value of \(ax^2+bx+c\) is ycoordinate of vertex. Attachment:
Math_cg_20 (1).png [ 18.71 KiB  Viewed 20202 times ]
Examples:Expression \(5x^210x+20\) reaches its minimum when \(x=\frac{b}{2a}=\frac{10}{2*5}=1\), so minimum value is \(5x^210x+20=5*1^210*1+20=15\). Expression \(5x^210x+20\) reaches its maximum when \(x=\frac{b}{2a}=\frac{10}{2*(5)}=1\), so maximum value is \(5x^210x+20=5*(1)^210*(1)+20=25\). Back to the original question:In a certain company, the formula for maximizing profits is P = 25x^2 + 7500x, where P is profit and x is the number of machines the company operates in its factory. What value for x will maximize P? A) 10 B) 50 C) 150 D) 200 E) 300 \(P=25x^2+7500x\) reaches its maximum (as \(a=25<0\)) when \(x=\frac{b}{2a}=\frac{7500}{2*(25)}=150\). Answer: C. As for the minimum value of P: minimum value of P is not limited (x increase to +infinity > P decreases to infinity). Hope it helps.
_________________




Manager
Joined: 30 May 2010
Posts: 157

Re: Finding Max or Min value of equation
[#permalink]
Show Tags
15 Aug 2010, 17:11
If you understand Calculus, this is the easy way of solving it. Nobody else mentioned, but you also need to take the second derivative.
\(f(x) = 25x^2 + 7500x\)
\(f'(x) = 50x + 7500\)
\(50x + 7500 = 0\) \(7500 = 50x\) \(x = 150\)
\(f''(x) = 50\)
If the second derivative is negative, it is a maximum. If it is positive, it is a minimum.




Manager
Joined: 27 May 2008
Posts: 105

Re: Finding Max or Min value of equation
[#permalink]
Show Tags
09 Feb 2009, 03:10
wcgmatclub wrote: This one is from IntegratedLearning. In a certain company, the formula for maximizing profits is P = 25×2 + 7500x, where P is profit and x is the number of machines the company operates in its factory. What value for x will maximize P?
A) 10 B) 50 C) 150 D) 200 E) 300
OC is C
Here's what I did: P=25x2+7500X => P=50+7500X, to max. P, we need to max X. So I picked E, why isn't that correct?
The OA states: To find a maximum or minimum value of an equation with an exponent in it, you take the derivative of the equation, set it to zero, and solve. I don't really get what that means. So whoever solves it, could you plz post explanation of what the above sentence mean as well? Pertaining to your question, What value for x will maximize P? if you get the derivative of P = 25×2 + 7500x then the equation will bcome, 50X + 7500 = 0 ( equating to 0 to get max value) X = 150



Intern
Joined: 15 Jan 2009
Posts: 43
Schools: Marshall '11

Re: Finding Max or Min value of equation
[#permalink]
Show Tags
09 Feb 2009, 12:26
Why does P needs to be set to 0 in order to get max value for P. I'm totally lost here. P = 25×2 + 7500x If x=150, then P=1124950 If x=300, then P=2249950 2249950>1124950, so P is greater when x=300 vs. x=150.
Why is x=150 the correct answer?



VP
Joined: 07 Nov 2007
Posts: 1086
Location: New York

Re: Finding Max or Min value of equation
[#permalink]
Show Tags
09 Feb 2009, 13:12
wcgmatclub wrote: Why does P needs to be set to 0 in order to get max value for P. I'm totally lost here. P = 25×2 + 7500x If x=150, then P=1124950 If x=300, then P=2249950 2249950>1124950, so P is greater when x=300 vs. x=150.
Why is x=150 the correct answer? Not p=0 its dP/dx =firt derviative of P with respect to x dP/dx =0 = 50x +7500 =0 x=150 Please not that we treated x2= x^2 (please use exponential symbol for that) P = 25×^2 + 7500x If x=150, then P=562500 If x=300, then P=0 Do you have knowledge of calculus (Derviatives)..?.



Intern
Joined: 15 Jan 2009
Posts: 43
Schools: Marshall '11

Re: Finding Max or Min value of equation
[#permalink]
Show Tags
09 Feb 2009, 15:22
x2suresh wrote: wcgmatclub wrote: Why does P needs to be set to 0 in order to get max value for P. I'm totally lost here. P = 25×2 + 7500x If x=150, then P=1124950 If x=300, then P=2249950 2249950>1124950, so P is greater when x=300 vs. x=150.
Why is x=150 the correct answer?
Not p=0 its dP/dx =firt derviative of P with respect to x dP/dx =0 = 50x +7500 =0 x=150 Please not that we treated x2= x^2 (please use exponential symbol for that)
P = 25×^2 + 7500x
If x=150, then P=562500 If x=300, then P=0
Do you have knowledge of calculus (Derviatives)..?. I took calculus a LONG time ago, so I prob. forgot all of it already. I thought GMAT doesn't test calculus concepts? Why is this "GMAT" type problem requires calculus to solve?



Intern
Status: Simply  Chasing GMAT
Joined: 04 May 2010
Posts: 18
Location: United Kingdom
Concentration: International Business, Entrepreneurship
GMAT Date: 01302012
GPA: 3
WE: Consulting (Computer Software)

Re: Finding Max or Min value of equation
[#permalink]
Show Tags
13 Aug 2010, 22:12
Thanks for this post to have a recall of calculus..."What value for x will maximize P" Need more clarification on minimum value....
what if require the value of x when profit is lowest???
My understanding is that it comes from the formula for minimize profit. This will be given in question stem... such as this is present here... "the formula for maximizing profits is P = 25x2 + 7500x"....
Please put the correct understanding here.



Intern
Status: Simply  Chasing GMAT
Joined: 04 May 2010
Posts: 18
Location: United Kingdom
Concentration: International Business, Entrepreneurship
GMAT Date: 01302012
GPA: 3
WE: Consulting (Computer Software)

Re: Finding Max or Min value of equation
[#permalink]
Show Tags
15 Aug 2010, 19:17
Hey Bro, Thanks a lot..."If the second derivative is negative, it is a maximum. If it is positive, it is a minimum. "...
I was looking for this only....its really long long time I left calculus so was trying some lead...Kudos Given..



Manager
Joined: 24 Apr 2013
Posts: 50
Location: United States

Re: In a certain company, the formula for maximizing profits is
[#permalink]
Show Tags
11 Sep 2013, 12:18
Hello I had two points to share 1) Please confirm that I understand the equation in the question correctly : P=(25)(2)+ (7500x). If I'm reading it correctly, then how does this equation turn into 50x+7500x? 2) How testable is this content on actual GMAT test? because I looked into the official guide testable topics as well as Manhattan books and it seems that calculus is not one of the topics. Where can I find this content to study?



Director
Joined: 03 Feb 2013
Posts: 838
Location: India
Concentration: Operations, Strategy
GPA: 3.88
WE: Engineering (Computer Software)

Re: In a certain company, the formula for maximizing profits is
[#permalink]
Show Tags
25 Jan 2014, 23:52
wcgc wrote: In a certain company, the formula for maximizing profits is P = 25×2 + 7500x, where P is profit and x is the number of machines the company operates in its factory. What value for x will maximize P? A) 10 B) 50 C) 150 D) 200 E) 300 This one is from IntegratedLearning. Here's what I did: P=25x2+7500X => P=50+7500X, to max. P, we need to max X. So I picked E, why isn't that correct?
The OA states: To find a maximum or minimum value of an equation with an exponent in it, you take the derivative of the equation, set it to zero, and solve. I don't really get what that means. So whoever solves it, could you plz post explanation of what the above sentence mean as well? People, please write the equation properly. I was wondering how p = 50 + 7500x we can solve this question for 5 mins It is simply as increasing functions. Please edit the question and write P = \(25 * x^2\) + \(7500x\) We can solve the above problem by calculas Maxima and minima or by using the perfect score approach.



Math Expert
Joined: 02 Sep 2009
Posts: 62499

Re: In a certain company, the formula for maximizing profits is
[#permalink]
Show Tags
26 Jan 2014, 03:22
kinjiGC wrote: wcgc wrote: In a certain company, the formula for maximizing profits is P = 25×2 + 7500x, where P is profit and x is the number of machines the company operates in its factory. What value for x will maximize P? A) 10 B) 50 C) 150 D) 200 E) 300 This one is from IntegratedLearning. Here's what I did: P=25x2+7500X => P=50+7500X, to max. P, we need to max X. So I picked E, why isn't that correct?
The OA states: To find a maximum or minimum value of an equation with an exponent in it, you take the derivative of the equation, set it to zero, and solve. I don't really get what that means. So whoever solves it, could you plz post explanation of what the above sentence mean as well? People, please write the equation properly. I was wondering how p = 50 + 7500x we can solve this question for 5 mins It is simply as increasing functions. Please edit the question and write P = \(25 * x^2\) + \(7500x\) We can solve the above problem by calculas Maxima and minima or by using the perfect score approach. ______ Done. Thank you.
_________________



Manager
Joined: 19 Nov 2012
Posts: 198
Concentration: Marketing, Social Entrepreneurship

Re: In a certain company, the formula for maximizing profits is
[#permalink]
Show Tags
09 Feb 2014, 18:08
If you're not comfortable with calculus, here is how I would do it.
Recognize that 25 is a factor of 7500. If we take this out, we have two parts to the equation:
X^2 & 300X
One part of the equation brings our value down, whereas the other part brings our value up. At this point, we can test the numbers in the answer choice. Notice that they are very straight forward to square, and multiplication by 300 is very easy.
A) 10  3000  100 = 2900 B) 50  15000  2500 = 12500 C) 150  45000  22500 = 22500 D) 200  60000  40000 = 20000 E) 300  recognize that this is zero
Therefore, answer is 150.
If you are able to spot the 25 in 7500, you can go through this process in and around 2 minutes. Also, if you tested C or D first, you'd recognize that A and B would never be able to reach their output, and spotting that E makes the equation equal to 0, there is only 2 numbers you would need to test.
Hope this helps those  like myself  who haven't thought about calculus for over half a decade.



Manager
Joined: 25 Mar 2013
Posts: 218
Location: United States
Concentration: Entrepreneurship, Marketing
GPA: 3.5

Re: In a certain company, the formula for maximizing profits is
[#permalink]
Show Tags
15 Feb 2014, 22:02
My approach is just plugging , X = 150 , where p = 562500 X = 300 , p = 0 , x = 200 = 500000 so C



Manager
Joined: 03 Jan 2015
Posts: 57
Concentration: Strategy, Marketing
WE: Research (Pharmaceuticals and Biotech)

Re: In a certain company, the formula for maximizing profits is
[#permalink]
Show Tags
30 Jan 2015, 22:48
In a certain company, the formula for maximizing profits is P = 25x^2 + 7500x,, where P is profit and x is the number of machines the company operates in its factory. What value for x will maximize P?
A) 10 B) 50 C) 150 D) 200 E) 300
SOLUTION:
P = 25x^2 + 7500x P = 25x (x + 300)  (I)
Plug in answer choices in (I):
A) 25*10 (290) B) 25*10*5 (250) C) 25*10*15 (150) D) 25*10*20 (100) E) 25*10*30 (0)
Dividing AE by 250: A) 290 B)1250 C)2250 D)2000 E) 0
ANSWER: C



Current Student
Joined: 12 Aug 2015
Posts: 2535

Re: In a certain company, the formula for maximizing profits is
[#permalink]
Show Tags
07 Mar 2016, 07:21
wcgc wrote: In a certain company, the formula for maximizing profits is P = 25x^2 + 7500x,, where P is profit and x is the number of machines the company operates in its factory. What value for x will maximize P? A) 10 B) 50 C) 150 D) 200 E) 300 This one is from IntegratedLearning. Here's what I did: P=25x2+7500X => P=50+7500X, to max. P, we need to max X. So I picked E, why isn't that correct?
The OA states: To find a maximum or minimum value of an equation with an exponent in it, you take the derivative of the equation, set it to zero, and solve. I don't really get what that means. So whoever solves it, could you plz post explanation of what the above sentence mean as well? Firstly I am not sure derivative will be useful in GMAT or not .. here is the solution though D(p)/D(x) = 50x+7500 Now the general rule is to equate the derivative to zero to get the values of any maxima or minima => 50x=7500 => x=150 Kudos if you like my solution ..It helps..
_________________



CEO
Joined: 20 Mar 2014
Posts: 2542
Concentration: Finance, Strategy
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)

In a certain company, the formula for maximizing profits is
[#permalink]
Show Tags
07 Mar 2016, 07:33
Chiragjordan wrote: wcgc wrote: In a certain company, the formula for maximizing profits is P = 25x^2 + 7500x,, where P is profit and x is the number of machines the company operates in its factory. What value for x will maximize P? A) 10 B) 50 C) 150 D) 200 E) 300 This one is from IntegratedLearning. Here's what I did: P=25x2+7500X => P=50+7500X, to max. P, we need to max X. So I picked E, why isn't that correct?
The OA states: To find a maximum or minimum value of an equation with an exponent in it, you take the derivative of the equation, set it to zero, and solve. I don't really get what that means. So whoever solves it, could you plz post explanation of what the above sentence mean as well? Firstly I am not sure derivative will be useful in GMAT or not .. here is the solution though D(p)/D(x) = 50x+7500 Now the general rule is to equate the derivative to zero to get the values of any maxima or minima => 50x=7500 => x=150 Kudos if you like my solution ..It helps.. Differentiation is not required for GMAT, but if you know it , then there is no harm in applying it. But for the sake of people who are not aware of differentiation, for all these max/min question involving quadratic equations, the best strategy is to come up with perfect squares of the form \((a \pm b)^2\) and then maximize or minimize by remembering the \((a \pm b)^2 \geq 0\) as shown below: P = \(25x^2 + 7500x\) > P = \(25 (x^2300x)\) = \(25 (x^22*150*x+150^2  150^2)\)= \(25 (x^22*150*x+150^2)+25*150^2\)= \(25 (x150)^2 + 25*150^2\) = a negative quantity + positive quantity. Thus to maximize P, you need to minimize the perfect square \((x150)^2\) and the minimum value of a perfect square = 0 > For P to be maximized , \((x150)^2 =0\) > \(x=150\). Any other value of x will only reduce PHence C is the correct answer. Hope this helps.



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2801

Re: In a certain company, the formula for maximizing profits is
[#permalink]
Show Tags
14 Jul 2017, 09:42
wcgc wrote: In a certain company, the formula for maximizing profits is P = 25x^2 + 7500x,, where P is profit and x is the number of machines the company operates in its factory. What value for x will maximize P?
A) 10 B) 50 C) 150 D) 200 E) 300 First we can see that the graph of P = 25x^2 + 7500x is a downopening parabola. This means that the vertex of the parabola will be at the maximum value. We find the xcoordinate of the vertex using the following equation: x = b/(2a) x = 7500/(2 x 25) = 7500/50 = 150 The maximum profit will be realized when 150 machines are sold. Answer: C
_________________
5star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews
If you find one of my posts helpful, please take a moment to click on the "Kudos" button.



Senior Manager
Joined: 28 Jun 2015
Posts: 272
Concentration: Finance
GPA: 3.5

In a certain company, the formula for maximizing profits is
[#permalink]
Show Tags
14 Jul 2017, 10:00
wcgc wrote: In a certain company, the formula for maximizing profits is P = 25x^2 + 7500x,, where P is profit and x is the number of machines the company operates in its factory. What value for x will maximize P? A) 10 B) 50 C) 150 D) 200 E) 300 This one is from IntegratedLearning. Here's what I did: P=25x2+7500X => P=50+7500X, to max. P, we need to max X. So I picked E, why isn't that correct?
The OA states: To find a maximum or minimum value of an equation with an exponent in it, you take the derivative of the equation, set it to zero, and solve. I don't really get what that means. So whoever solves it, could you plz post explanation of what the above sentence mean as well? Here's how I did it... \(P = 25x^2 + 7500x = 25x(x  300)\) For \(P\) to be positive, \(x\) must be less than 300, so option E is out. (D): \(25(200) (100) = 500,000\) (C): \(25(150) (150) = 562,500\) (B): \(25(50) (250) = 312,500\) (A): \(25(10) (290) = 72,500\) \(P\) is maximum when \(x = 150\), so option C. Using calculus: \(25x^2 + 7500x\) is maximum when \(\frac{d}{dx}(25x^2 + 7500x) = 0\). \(\frac{d}{dx}(25x^2 + 7500x) = 0\) \(50x + 7500 = 0\) \(50x = 7500\) \(x = \frac{7500}{50} = 150.\)
_________________
I used to think the brain was the most important organ. Then I thought, look what’s telling me that.



Intern
Joined: 04 Nov 2018
Posts: 13
Concentration: Finance, General Management
GMAT 1: 700 Q49 V35 GMAT 2: 710 Q48 V40
GPA: 3.83
WE: Sales (Retail)

Re: In a certain company, the formula for maximizing profits is
[#permalink]
Show Tags
24 Mar 2019, 00:37
What I did was first solve for what values minimise the profit. By solving for =0, we find that x is either 0 or 300. BUT, these values minimise the profit, however we want to maximise. So with a quick look at the possible answers, the "middle" value between these two extremes that minimise the profit, will maximise it. Not very mathematically correct, I guess, but works. No?
_________________
 GMAT Prep #1 CAT (Apr 2019) : 640 (Q48, V30)  GMAT Prep #1 CAT (early Oct 2019  post hiatus) : 680 (Q48, V34)  MGMAT CAT #1 (mid Oct 2019) : 640 (Q44, V34)  MGMAT CAT #2 (midtolate Oct 2019) : 610 (Q40, V34)  GMAT Prep #2 CAT (late Oct 2019) : 720 (Q49, V40)  MGMAT CAT #3 (early Nov 2019) : 660 (Q42, V38)  GMAT 1 : 700 (Q49, V35)
Still not there.
If you're reading this, we've got this.




Re: In a certain company, the formula for maximizing profits is
[#permalink]
24 Mar 2019, 00:37






