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In a certain company, the formula for maximizing profits is
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Updated on: 26 Jan 2014, 04:21
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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?
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Originally posted by wcgc on 09 Feb 2009, 03:12.
Last edited by Bunuel on 26 Jan 2014, 04:21, edited 1 time in total.
Edited the question.




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Re: Finding Max or Min value of equation
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14 Aug 2010, 07:32
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 20032 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.




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Re: Finding Max or Min value of equation
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15 Aug 2010, 18:11
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.




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Re: Finding Max or Min value of equation
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09 Feb 2009, 04:10
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



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Re: Finding Max or Min value of equation
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09 Feb 2009, 13:26
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?



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Re: Finding Max or Min value of equation
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09 Feb 2009, 14:12
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)..?.



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Re: Finding Max or Min value of equation
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09 Feb 2009, 16:22
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?



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Re: Finding Max or Min value of equation
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13 Aug 2010, 23:12
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.



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Re: Finding Max or Min value of equation
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15 Aug 2010, 20:17
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..



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Re: In a certain company, the formula for maximizing profits is
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11 Sep 2013, 13:18
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?



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Re: In a certain company, the formula for maximizing profits is
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26 Jan 2014, 00:52
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.



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Re: In a certain company, the formula for maximizing profits is
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26 Jan 2014, 04:22
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.



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Re: In a certain company, the formula for maximizing profits is
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09 Feb 2014, 19:08
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.



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Re: In a certain company, the formula for maximizing profits is
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15 Feb 2014, 23:02
My approach is just plugging , X = 150 , where p = 562500 X = 300 , p = 0 , x = 200 = 500000 so C



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Re: In a certain company, the formula for maximizing profits is
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30 Jan 2015, 23:48
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



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Re: In a certain company, the formula for maximizing profits is
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07 Mar 2016, 08:21
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..
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In a certain company, the formula for maximizing profits is
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07 Mar 2016, 08:33
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.



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Re: In a certain company, the formula for maximizing profits is
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14 Jul 2017, 10:42
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
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In a certain company, the formula for maximizing profits is
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14 Jul 2017, 11:00
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 \(\frac{d}{dx}(25x^2 + 7500x) = 0\). \(\frac{d}{dx}(25x^2 + 7500x) = 0\) \(50x + 7500 = 0\) \(50x = 7500\) \(x = \frac{7500}{50} = 150.\)
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Re: In a certain company, the formula for maximizing profits is
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24 Mar 2019, 01:37
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?
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