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The revenue, R, from selling a certain type of flower varies with the

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The revenue, R, from selling a certain type of flower varies with the  [#permalink]

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New post 02 Jan 2018, 23:45
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A
B
C
D
E

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  5% (low)

Question Stats:

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

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Re: The revenue, R, from selling a certain type of flower varies with the  [#permalink]

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New post 03 Jan 2018, 03:42
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.
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New post 03 Jan 2018, 03:46
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.
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Re: The revenue, R, from selling a certain type of flower varies with the  [#permalink]

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New post 03 Jan 2018, 04:16
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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


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
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Re: The revenue, R, from selling a certain type of flower varies with the  [#permalink]

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New post 03 Jan 2018, 09:41
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.
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Re: The revenue, R, from selling a certain type of flower varies with the  [#permalink]

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New post 01 Dec 2018, 10:05
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.
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Re: The revenue, R, from selling a certain type of flower varies with the  [#permalink]

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New post 06 Aug 2019, 17:02
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
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Re: The revenue, R, from selling a certain type of flower varies with the   [#permalink] 06 Aug 2019, 17:02
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