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Bunuel
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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Bunuel
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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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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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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