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Iteru [2.4K]
2 years ago
10

Kristin owns a bakery called Kristin's Cakes n' Such and is considering lowering the price of her cakes. Kristin polls her custo

mers and determines that she can sell 100 cakes each week when she chargers $25 each. She also discovers that for every $1 decrease in the price of the cake, she will sell 5 more cakes. At what price should Kristin sell her cakes in order to reach he maximum potential revenue?
Mathematics
1 answer:
azamat2 years ago
7 0
The total revenue that is gained from the sales of the cakes is determined by multiplying the number of cakes by the price. If we let x be the number of $1 that should be deducted from the price and y be the total revenue,
                               y = (25 - x)(100 + 5x)
Simplifying, 
                              y = 2500 + 25x - 5x²
We get the value of x that will give us the maximum revenue by differentiating the equation and equating the differential to zero.
                             dy/dx = 0 = 25 - 10x
The value of x is 2.5. 
 The price of the cake should be 25 - 2.5 = 22.5. 
Thus, the price of the cake that will give the maximum potential revenue is $22.5. 
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Drag each tile to the correct box.
Umnica [9.8K]

Answer:

Emma ⇒ IQR = 3.5

Sally ⇒ IQR = 3

Olivia ⇒ IQR = 2.5

Emma - Sally - Olivia

Step-by-step explanation:

Emma's Scores:

<u>6 , 7 , 8 , 8 , </u><u>9 , 9</u><u> , 9 , 10 , 10 , </u><u>11</u> , <u>11</u><u> , 11 , 11 , 11 , </u><u>12 , 13</u><u> , 13 , 13 , 13 , 14</u>

LQ = (9 + 9) ÷ 2 = 9

UQ = (12 + 13) ÷ 2 = 12.5

IQR = 12.5 - 9 = 3.5

Sally's Scores:

<u>6 , 6 , 6 , 7 , </u><u>7 , 8</u><u> , 8 , 8 , 9 , </u><u>9</u> , <u>10</u><u> , 10 , 10 , 10 , 10</u><u> , 10 , 11</u><u> , 11 , 12 , 12 , 14</u>

LQ = (7 + 8) ÷ 2 = 7.5

UQ = (10 + 11) ÷ 2 = 10.5

IQR = 10.5 - 7.5 = 3

Olivia's Scores:

<u>6 , 6 , 7 , 7 , </u><u>7 , 8</u><u> , 8 , 8 , 8 , </u><u>9</u> , <u>9</u><u> , 9 , 9 , 9 , </u><u>10 , 10</u><u> , 12 , 12 , 13 , 14</u>

LQ = (7 + 8) ÷ 2 = 7.5

UQ = (10 + 10) ÷ 2 = 10

IQR = 10 - 7.5 = 2.5

Emma - Sally - Olivia

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