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Sagot :
To find the price that maximizes the store's revenue, we need to follow a systematic approach. The revenue (R) can be expressed as:
[tex]\[ R = p \times \text{(number of backpacks sold)} \][/tex]
Given the expression for the number of backpacks sold per day:
[tex]\[ \text{Number of backpacks sold} = -2p + 50 \][/tex]
Substituting this into the revenue expression, we get:
[tex]\[ R = p \times (-2p + 50) \][/tex]
We need to calculate this for each price [tex]\( p \)[/tex] in the range from \[tex]$9 to \$[/tex]15 and find which price gives the highest revenue.
1. Set up a table to calculate revenue for each price:
| Price [tex]\(p\)[/tex] (in dollars) | Number of backpacks sold ([tex]\(-2p + 50\)[/tex]) | Revenue [tex]\(R = p \times (-2p + 50)\)[/tex] |
|---------------------------|------------------------------------------|-------------------------------------|
| 9 | [tex]\(-2(9) + 50 = 32\)[/tex] | [tex]\(9 \times 32 = 288\)[/tex] |
| 10 | [tex]\(-2(10) + 50 = 30\)[/tex] | [tex]\(10 \times 30 = 300\)[/tex] |
| 11 | [tex]\(-2(11) + 50 = 28\)[/tex] | [tex]\(11 \times 28 = 308\)[/tex] |
| 12 | [tex]\(-2(12) + 50 = 26\)[/tex] | [tex]\(12 \times 26 = 312\)[/tex] |
| 13 | [tex]\(-2(13) + 50 = 24\)[/tex] | [tex]\(13 \times 24 = 312\)[/tex] |
| 14 | [tex]\(-2(14) + 50 = 22\)[/tex] | [tex]\(14 \times 22 = 308\)[/tex] |
| 15 | [tex]\(-2(15) + 50 = 20\)[/tex] | [tex]\(15 \times 20 = 300\)[/tex] |
2. Analyze the calculated revenues:
- At [tex]\( p = 9 \)[/tex], the revenue is [tex]\( 288 \)[/tex].
- At [tex]\( p = 10 \)[/tex], the revenue is [tex]\( 300 \)[/tex].
- At [tex]\( p = 11 \)[/tex], the revenue is [tex]\( 308 \)[/tex].
- At [tex]\( p = 12 \)[/tex], the revenue is [tex]\( 312 \)[/tex].
- At [tex]\( p = 13 \)[/tex], the revenue is [tex]\( 312 \)[/tex].
- At [tex]\( p = 14 \)[/tex], the revenue is [tex]\( 308 \)[/tex].
- At [tex]\( p = 15 \)[/tex], the revenue is [tex]\( 300 \)[/tex].
3. Determine the maximum revenue and corresponding price:
From the table, we observe that the highest revenue is [tex]\( \$312 \)[/tex]. This maximum revenue occurs at two different prices: [tex]\( \$12 \)[/tex] and [tex]\( \$13 \)[/tex]. However, since the problem specifically asks for a single price, we may choose either.
Thus, the price that gives the store the maximum revenue is \[tex]$12 (or \$[/tex]13), and the maximum revenue is \$312.
[tex]\[ R = p \times \text{(number of backpacks sold)} \][/tex]
Given the expression for the number of backpacks sold per day:
[tex]\[ \text{Number of backpacks sold} = -2p + 50 \][/tex]
Substituting this into the revenue expression, we get:
[tex]\[ R = p \times (-2p + 50) \][/tex]
We need to calculate this for each price [tex]\( p \)[/tex] in the range from \[tex]$9 to \$[/tex]15 and find which price gives the highest revenue.
1. Set up a table to calculate revenue for each price:
| Price [tex]\(p\)[/tex] (in dollars) | Number of backpacks sold ([tex]\(-2p + 50\)[/tex]) | Revenue [tex]\(R = p \times (-2p + 50)\)[/tex] |
|---------------------------|------------------------------------------|-------------------------------------|
| 9 | [tex]\(-2(9) + 50 = 32\)[/tex] | [tex]\(9 \times 32 = 288\)[/tex] |
| 10 | [tex]\(-2(10) + 50 = 30\)[/tex] | [tex]\(10 \times 30 = 300\)[/tex] |
| 11 | [tex]\(-2(11) + 50 = 28\)[/tex] | [tex]\(11 \times 28 = 308\)[/tex] |
| 12 | [tex]\(-2(12) + 50 = 26\)[/tex] | [tex]\(12 \times 26 = 312\)[/tex] |
| 13 | [tex]\(-2(13) + 50 = 24\)[/tex] | [tex]\(13 \times 24 = 312\)[/tex] |
| 14 | [tex]\(-2(14) + 50 = 22\)[/tex] | [tex]\(14 \times 22 = 308\)[/tex] |
| 15 | [tex]\(-2(15) + 50 = 20\)[/tex] | [tex]\(15 \times 20 = 300\)[/tex] |
2. Analyze the calculated revenues:
- At [tex]\( p = 9 \)[/tex], the revenue is [tex]\( 288 \)[/tex].
- At [tex]\( p = 10 \)[/tex], the revenue is [tex]\( 300 \)[/tex].
- At [tex]\( p = 11 \)[/tex], the revenue is [tex]\( 308 \)[/tex].
- At [tex]\( p = 12 \)[/tex], the revenue is [tex]\( 312 \)[/tex].
- At [tex]\( p = 13 \)[/tex], the revenue is [tex]\( 312 \)[/tex].
- At [tex]\( p = 14 \)[/tex], the revenue is [tex]\( 308 \)[/tex].
- At [tex]\( p = 15 \)[/tex], the revenue is [tex]\( 300 \)[/tex].
3. Determine the maximum revenue and corresponding price:
From the table, we observe that the highest revenue is [tex]\( \$312 \)[/tex]. This maximum revenue occurs at two different prices: [tex]\( \$12 \)[/tex] and [tex]\( \$13 \)[/tex]. However, since the problem specifically asks for a single price, we may choose either.
Thus, the price that gives the store the maximum revenue is \[tex]$12 (or \$[/tex]13), and the maximum revenue is \$312.
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