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# A shoe store sells a new model of running shoe for $32. At that price SORT BY: Tags: Show Tags Hide Tags Math Expert Joined: 02 Sep 2009 Posts: 93417 Own Kudos [?]: 626032 [24] Given Kudos: 81940 Most Helpful Reply Math Expert Joined: 02 Sep 2009 Posts: 93417 Own Kudos [?]: 626032 [1] Given Kudos: 81940 General Discussion Manager Joined: 20 Dec 2019 Posts: 105 Own Kudos [?]: 100 [3] Given Kudos: 74 GMATWhiz Representative Joined: 23 May 2022 Posts: 640 Own Kudos [?]: 438 [3] Given Kudos: 6 Location: India GMAT 1: 760 Q51 V40 Re: A shoe store sells a new model of running shoe for$32. At that price [#permalink]
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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 Quant Chat Moderator Joined: 22 Dec 2016 Posts: 3139 Own Kudos [?]: 4466 [2] Given Kudos: 1856 Location: India Concentration: Strategy, Leadership Re: A shoe store sells a new model of running shoe for$32. At that price [#permalink]
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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. Attachments Screenshot 2022-09-16 110103.png [ 226.28 KiB | Viewed 3651 times ] Intern Joined: 11 Feb 2024 Posts: 1 Own Kudos [?]: 0 [0] Given Kudos: 6 A shoe store sells a new model of running shoe for$32. At that price [#permalink]
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]
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