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Math Expert V
Joined: 02 Sep 2009
Posts: 58445
The revenue, R, from selling a certain type of flower varies with the  [#permalink]

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Difficulty:   5% (low)

Question Stats: 85% (01:12) correct 15% (01:07) wrong based on 122 sessions

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The revenue, R, from selling a certain type of flower varies with the price per flower according to the equation $$R = –40(p – 3)^2 + 360$$. What price per flower, p, should be charged to achieve the highest revenue?

A. $36 B.$12
C. $3 D.$2
E. $1 _________________ Retired Moderator D Joined: 25 Feb 2013 Posts: 1178 Location: India GPA: 3.82 Re: The revenue, R, from selling a certain type of flower varies with the [#permalink] Show Tags Bunuel wrote: The revenue, R, from selling a certain type of flower varies with the price per flower according to the equation R = –40(p – 3)2+ 360. What price per flower, p, should be charged to achieve the highest revenue? A.$36
B. $12 C.$3
D. $2 E.$1

Hi Bunuel

Kindly confirm whether the highlighted part is $$(p-3)^2$$ or is it simply $$(p-3)*2$$ because in either case answer will be different.
Math Expert V
Joined: 02 Sep 2009
Posts: 58445
Re: The revenue, R, from selling a certain type of flower varies with the  [#permalink]

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niks18 wrote:
Bunuel wrote:
The revenue, R, from selling a certain type of flower varies with the price per flower according to the equation R = –40(p – 3)2+ 360. What price per flower, p, should be charged to achieve the highest revenue?

A. $36 B.$12
C. $3 D.$2
E. $1 Hi Bunuel Kindly confirm whether the highlighted part is $$(p-3)^2$$ or is it simply $$(p-3)*2$$ because in either case answer will be different. It's $$R = –40(p – 3)^2 + 360$$. Edited. Thank you. _________________ Retired Moderator D Joined: 25 Feb 2013 Posts: 1178 Location: India GPA: 3.82 Re: The revenue, R, from selling a certain type of flower varies with the [#permalink] Show Tags 1 Bunuel wrote: The revenue, R, from selling a certain type of flower varies with the price per flower according to the equation $$R = –40(p – 3)^2 + 360$$. What price per flower, p, should be charged to achieve the highest revenue? A.$36
B. $12 C.$3
D. $2 E.$1

To maximize $$R$$ we need to minimize $$-40(p-3)^2$$ which is a negative value irrespective of $$p$$

Hence at $$p=3$$; $$-40(p-3)^2=0$$ hence $$R=360$$

Option C
Ask GMAT Experts Forum Moderator V
Status: Preparing for GMAT
Joined: 25 Nov 2015
Posts: 1044
Location: India
GPA: 3.64
Re: The revenue, R, from selling a certain type of flower varies with the  [#permalink]

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Bunuel wrote:
The revenue, R, from selling a certain type of flower varies with the price per flower according to the equation $$R = –40(p – 3)^2 + 360$$. What price per flower, p, should be charged to achieve the highest revenue?

A. $36 B.$12
C. $3 D.$2
E. $1 $$R = –40(p – 3)^2 + 360$$ To achieve the maximum revenue, we must decrease the term $$–40(p – 3)^2$$, which becomes zero when p=3. Answer C. _________________ Please give kudos, if you like my post When the going gets tough, the tough gets going... Manager  S Joined: 12 Apr 2017 Posts: 139 Location: United States Concentration: Finance, Operations GPA: 3.1 Re: The revenue, R, from selling a certain type of flower varies with the [#permalink] Show Tags Great question! I had to stop and think and not just pick an answer. Answer is C, just plugging in the answers you get C at$360, which is the max.

A & B, result in negative values > than 360.
Intern  B
Joined: 24 Aug 2017
Posts: 21
Schools: INSEAD, CEIBS '22
GPA: 3.67
Re: The revenue, R, from selling a certain type of flower varies with the  [#permalink]

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Another approach (but a bit longer)

We know that the for y=ax^2 + bx +c, the vertex will be at a point (x,y) where x = -b/2a

Expand to -40p^2 + 240 p + 720
- 240/ 2(-40) = 3 Re: The revenue, R, from selling a certain type of flower varies with the   [#permalink] 06 Aug 2019, 17:02
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