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01 Apr 2024, 08:40
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A shoe store sells a new model of running shoe for $32. At that price, the store sells 80 pairs each week. The manager however estimated that the store will sell 20 additional pairs of shoes each week for every $2 reduction in the price. According to these estimations, what is the maximum amount by which the store can increase its weekly revenue from the shoes?
A. 1,440
B. 1,600
C. 2,460
D. 3,920
E. 4,000
A. 1,440
B. 1,600
C. 2,460
D. 3,920
E. 4,000
01 Apr 2024, 08:40
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Official Solution:
A shoe store sells a new model of running shoe for $32. At that price, the store sells 80 pairs each week. The manager however estimated that the store will sell 20 additional pairs of shoes each week for every $2 reduction in the price. According to these estimations, what is the maximum amount by which the store can increase its weekly revenue from the shoes?
A. 1,440
B. 1,600
C. 2,460
D. 3,920
E. 4,000
Assume that to achieve the maximum difference between the current and estimated revenue, the price should be reduced by $2 \(x\) times. Then the price of a pair of shoes would be \(32 - 2x\) dollars, and a total of \(80 + 20x\) pairs of shoes would be sold, generating revenue equal to \((32 - 2x)(80 + 20x)\) dollars.
We want to maximize \((32 - 2x)(80 + 20x)\). Let's work with the expression:
Complete the square for \(x^2 - 12 x - 64\):
Since the value of \(-40(x - 6)^2\) is 0 or negative, the maximum value of \(-40(x - 6)^2 + 4,000\) is \(4,000\), that is when \(-40(x - 6)^2\) is 0. Therefore, the maximum amount by which the store can increase its weekly revenue from the shoes is \(4,000 - 32*80 = 1,440\) dollars.
Answer: A
A shoe store sells a new model of running shoe for $32. At that price, the store sells 80 pairs each week. The manager however estimated that the store will sell 20 additional pairs of shoes each week for every $2 reduction in the price. According to these estimations, what is the maximum amount by which the store can increase its weekly revenue from the shoes?
A. 1,440
B. 1,600
C. 2,460
D. 3,920
E. 4,000
Assume that to achieve the maximum difference between the current and estimated revenue, the price should be reduced by $2 \(x\) times. Then the price of a pair of shoes would be \(32 - 2x\) dollars, and a total of \(80 + 20x\) pairs of shoes would be sold, generating revenue equal to \((32 - 2x)(80 + 20x)\) dollars.
We want to maximize \((32 - 2x)(80 + 20x)\). Let's work with the expression:
\((32 - 2x)(80 + 20x)=\)
\(=-40(x - 16)(4 + x)=\)
\(=-40(x^2 - 12x - 64)\)
\(=-40(x - 16)(4 + x)=\)
\(=-40(x^2 - 12x - 64)\)
Complete the square for \(x^2 - 12 x - 64\):
\(=-40(x^2 - 12x + 36 -36 - 64)\)
\(=-40[(x - 6)^2 - 100)]\)
\(=-40(x - 6)^2 + 4,000\)
\(=-40[(x - 6)^2 - 100)]\)
\(=-40(x - 6)^2 + 4,000\)
Since the value of \(-40(x - 6)^2\) is 0 or negative, the maximum value of \(-40(x - 6)^2 + 4,000\) is \(4,000\), that is when \(-40(x - 6)^2\) is 0. Therefore, the maximum amount by which the store can increase its weekly revenue from the shoes is \(4,000 - 32*80 = 1,440\) dollars.
Answer: A
06 Apr 2024, 10:49
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or after getting the the equation -40(x^2-12x-64) we can deffrentiate and get the d(40(x^2-12x-64))/dx =0
2x- 12=0
x=6 is the maxima point
so substituting in the above equation we get 4000 and we solve like above
just another method!
2x- 12=0
x=6 is the maxima point
so substituting in the above equation we get 4000 and we solve like above
just another method!
