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To determine the price that gives the store the maximum revenue and calculate what that maximum revenue is, we can use the given function that models the number of backpacks sold per day as a function of price, [tex]\( -2p + 50 \)[/tex].
First, let’s define the revenue function. Revenue, [tex]\( R \)[/tex], is given by the product of the price per backpack, [tex]\( p \)[/tex], and the number of backpacks sold, which is modeled by [tex]\( -2p + 50 \)[/tex]:
[tex]\[ R(p) = p \times (-2p + 50) = -2p^2 + 50p \][/tex]
Next, we need to evaluate the revenue function at specific prices: [tex]$9.00, $[/tex]12.00, [tex]$12.50, and $[/tex]15.00 to determine which gives the maximum revenue.
1. Evaluate at [tex]\( p = 9.00 \)[/tex]:
[tex]\[ R(9) = -2(9)^2 + 50(9) = -2(81) + 450 = -162 + 450 = 288 \][/tex]
2. Evaluate at [tex]\( p = 12.00 \)[/tex]:
[tex]\[ R(12) = -2(12)^2 + 50(12) = -2(144) + 600 = -288 + 600 = 312 \][/tex]
3. Evaluate at [tex]\( p = 12.50 \)[/tex]:
[tex]\[ R(12.5) = -2(12.5)^2 + 50(12.5) = -2(156.25) + 625 = -312.5 + 625 = 312.5 \][/tex]
4. Evaluate at [tex]\( p = 15.00 \)[/tex]:
[tex]\[ R(15) = -2(15)^2 + 50(15) = -2(225) + 750 = -450 + 750 = 300 \][/tex]
Comparing the revenues calculated:
[tex]\[ R(9.00) = 288 \][/tex]
[tex]\[ R(12.00) = 312 \][/tex]
[tex]\[ R(12.50) = 312.5 \][/tex]
[tex]\[ R(15.00) = 300 \][/tex]
The maximum revenue occurs at [tex]\( p = 12.50 \)[/tex], with the revenue being [tex]$312.50. Thus, the price that gives the store the maximum revenue is \( \$[/tex]12.50 \) per backpack, and the maximum revenue is [tex]\( \$312.50 \)[/tex].
First, let’s define the revenue function. Revenue, [tex]\( R \)[/tex], is given by the product of the price per backpack, [tex]\( p \)[/tex], and the number of backpacks sold, which is modeled by [tex]\( -2p + 50 \)[/tex]:
[tex]\[ R(p) = p \times (-2p + 50) = -2p^2 + 50p \][/tex]
Next, we need to evaluate the revenue function at specific prices: [tex]$9.00, $[/tex]12.00, [tex]$12.50, and $[/tex]15.00 to determine which gives the maximum revenue.
1. Evaluate at [tex]\( p = 9.00 \)[/tex]:
[tex]\[ R(9) = -2(9)^2 + 50(9) = -2(81) + 450 = -162 + 450 = 288 \][/tex]
2. Evaluate at [tex]\( p = 12.00 \)[/tex]:
[tex]\[ R(12) = -2(12)^2 + 50(12) = -2(144) + 600 = -288 + 600 = 312 \][/tex]
3. Evaluate at [tex]\( p = 12.50 \)[/tex]:
[tex]\[ R(12.5) = -2(12.5)^2 + 50(12.5) = -2(156.25) + 625 = -312.5 + 625 = 312.5 \][/tex]
4. Evaluate at [tex]\( p = 15.00 \)[/tex]:
[tex]\[ R(15) = -2(15)^2 + 50(15) = -2(225) + 750 = -450 + 750 = 300 \][/tex]
Comparing the revenues calculated:
[tex]\[ R(9.00) = 288 \][/tex]
[tex]\[ R(12.00) = 312 \][/tex]
[tex]\[ R(12.50) = 312.5 \][/tex]
[tex]\[ R(15.00) = 300 \][/tex]
The maximum revenue occurs at [tex]\( p = 12.50 \)[/tex], with the revenue being [tex]$312.50. Thus, the price that gives the store the maximum revenue is \( \$[/tex]12.50 \) per backpack, and the maximum revenue is [tex]\( \$312.50 \)[/tex].
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