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A shoe store sells a new model of running shoe for $32. At that price

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Bunuel
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A shoe store sells a new model of running shoe for $32. At that price [#permalink] A shoe store sells a new model of running shoe for $32. At that price   14 Sep 2022, 07:18
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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. 1440
B. 1600
C. 2460
D. 3920
E. 4000


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Bunuel
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Re: A shoe store sells a new model of running shoe for $32. At that price [#permalink] A shoe store sells a new model of running shoe for $32. At that price   01 Apr 2024, 08:45
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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


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:

\((32 - 2x)(80 + 20x)=\)

\(=-40(x - 16)(4 + x)=\)

\(=-40(x^2 - 12x - 64)\)

Complete the square for \(x^2 - 12 x - 64\):

\(=-40(x^2 - 12x + 36 -36 - 64)\)

\(=-40[(x - 6)^2 - 100)]\)

\(=-40(x - 6)^2 + 4,000\)

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
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Re: A shoe store sells a new model of running shoe for $32. At that price [#permalink] A shoe store sells a new model of running shoe for $32. At that price   14 Sep 2022, 07:40
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Bunuel wrote:
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. 1440
B. 1600
C. 2460
D. 3920
E. 4000


Check twin question HERE.

 


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Present weekly revenue = 80 x $32 = $2560

So according to estimation, Sales will increase by 20 pairs for every $2 decrease in price
So the possibilities are
100 x $30 = $3000
120 x $28 = $3360
140 x $26 = ....
.
..
.
.
.
200 x $20 = $4000 (Hereafter it starts decreasing)

So the maximum amount by which the store can increase its weekly revenue = $4000 - $2560 = $1440

Ans is A
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Re: A shoe store sells a new model of running shoe for $32. At that price [#permalink] A shoe store sells a new model of running shoe for $32. At that price   15 Sep 2022, 01:14
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Bunuel wrote:
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. 1440
B. 1600
C. 2460
D. 3920
E. 4000

Solution:

  • When selling the shoe at $32, the store sells 80 pairs
  • Reducing the price by $2, increases the sale by 20
  • Let us assume that this reduction os $2 is done n times
  • So, the revenue, in that case, will be \((80+20n)(32-2n)\)
  • We need to maximise \((80+20n)(32-2n)\)
    \(⇒2560-160n+640n-40n^2\)
    \(⇒-40n^2+480n+2560\)
    \(⇒-40(n^2-12n-64)\)
    \(⇒-40(n^2-12n+6^2-6^2-64)\)
    \(⇒-40[(n-6)^2-36-64]\)
    \(⇒-40[(n-6)^2-100]\)
    \(⇒-40(n-6)^2+4000\)
  • We can infer that the maximum value of \(-40(n-6)^2+4000\)is 4000 i.e., when \(-40(n-6)^2=0\)
  • Increase in the revenue \(=4000-2560=$1440\)


Hence the right answer is Option A
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Re: A shoe store sells a new model of running shoe for $32. At that price [#permalink] A shoe store sells a new model of running shoe for $32. At that price   15 Sep 2022, 22:37
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Bunuel wrote:
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. 1440
B. 1600
C. 2460
D. 3920
E. 4000

 


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We are given that for every $2 decrease in price, 20 additional pairs of shoes are sold.

Current Revenue = 32 * 80 = $2560

Let's assume that we need to decrease the price by 2x to obtain maximum revenue.

At that price, the number of pair of shoes that will be sold is (100 + 20x)

Revenue = (100 + 2x) (32-2x)

We see that this is a quadratic equation, and a downward parabola. The maxima will lie at \(\frac{-b}{2a}\)

Upon solving x = 6

Revenue at x = 6

= (32 - 12)(80 + 120) = 4000

Difference = 4000 - 2560 = $1440

Option A

Detailed working is shown in the attached image.
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A shoe store sells a new model of running shoe for $32. At that price [#permalink] A shoe store sells a new model of running shoe for $32. At that price   10 May 2024, 08:00
Another way to solve:
Once you get to the quadratic equation (x^2−12x−64) 
Factor to: (x-16)(x+4)
Solve for x's: x = 16, x = -4
As is parabola, maximum point is between the two x-values, so (16+(-4))/2 = 6

Then continue as other solutions:
new revenue = (80+20(6))*(32-2(6)) = 4000
old revenue = 32*80 = 2560
4000 - 2560 = 1440
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A shoe store sells a new model of running shoe for $32. At that price [#permalink]
10 May 2024, 08:00
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