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In a certain company, the formula for maximizing profits is [#permalink]
09 Feb 2009, 02:12
7
This post was BOOKMARKED
00:00
A
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Difficulty:
55% (hard)
Question Stats:
50% (02:12) correct
50% (01:29) wrong based on 307 sessions
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?
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?
Re: Finding Max or Min value of equation [#permalink]
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
Re: Finding Max or Min value of equation [#permalink]
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.
Re: Finding Max or Min value of equation [#permalink]
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)..?. _________________
Your attitude determines your altitude Smiling wins more friends than frowning
Re: Finding Max or Min value of equation [#permalink]
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?
Re: Finding Max or Min value of equation [#permalink]
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. _________________
If you like my post, consider giving me a KUDOS. THANKS!!!
Re: Finding Max or Min value of equation [#permalink]
14 Aug 2010, 06:32
9
This post received KUDOS
Expert's post
8
This post was BOOKMARKED
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 y-coordinate of vertex, when \(a<0\), the parabola opens downward and maximum value of \(ax^2+bx+c\) is y-coordinate of vertex.
Attachment:
Math_cg_20 (1).png [ 18.71 KiB | Viewed 15473 times ]
Examples: Expression \(5x^2-10x+20\) reaches its minimum when \(x=-\frac{b}{2a}=-\frac{-10}{2*5}=1\), so minimum value is \(5x^2-10x+20=5*1^2-10*1+20=15\).
Expression \(-5x^2-10x+20\) reaches its maximum when \(x=-\frac{b}{2a}=-\frac{-10}{2*(-5)}=-1\), so maximum value is \(-5x^2-10x+20=-5*(-1)^2-10*(-1)+20=25\).
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).
Re: In a certain company, the formula for maximizing profits is [#permalink]
11 Sep 2013, 08:36
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Re: In a certain company, the formula for maximizing profits is [#permalink]
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? _________________
Re: In a certain company, the formula for maximizing profits is [#permalink]
25 Jan 2014, 23:52
1
This post received KUDOS
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?
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. _________________
Re: In a certain company, the formula for maximizing profits is [#permalink]
26 Jan 2014, 03:22
Expert's post
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?
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.
Re: In a certain company, the formula for maximizing profits is [#permalink]
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.
Re: In a certain company, the formula for maximizing profits is [#permalink]
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 A-E by 250: A) 290 B)1250 C)2250 D)2000 E) 0
ANSWER: C
gmatclubot
Re: In a certain company, the formula for maximizing profits is
[#permalink]
30 Jan 2015, 22:48
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