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

Bunuel

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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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01 Apr 2024, 08:45

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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:

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

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

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

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

General Discussion

ankitgoswami

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14 Sep 2022, 07:40

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Bunuel

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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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15 Sep 2022, 01:14

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Bunuel

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

gmatophobia

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Bunuel

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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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Screenshot 2022-09-16 110103.png [ 226.28 KiB | Viewed 10128 times ]

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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

sayan640

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15 Sep 2024, 03:54

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The store will sell 20 additional pairs of shoes each week for every $2 reduction in the price.
Lets say , this happens for 'c' times.

New revenue will be = 32*80 * (1 + c/4) * (1 - c/16)

Because increase in shoes = 20/80 = 1/4
Decrease in price = 2/32 = 1/16

Hence for maximising , we need to maximise = (1 + c/4) * (1 - c/16)
IF we differentiate , we get the value of c = 6

At c= 6 , the revenue will be maximum.
The maximum revenue will be = 32.80 . ( 1 + 6/4) * ( 1-6/16) = 4000

Initial revenue = 32 * 80 = 2560
Hence , the maximum by which we can increase = 4000 - 2560 = 1440

ans A.
Kindly check KarishmaB GMATinsight

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Hey Bunuel / KarishmaB / gmatophobia

Might be a beginner question - I'm facing difficulty understanding the solution from this step,

[ltr]
1. Why are we manipulating the quadratic equation, rather the solving it upfront?

2. Would we not reach the soln. by solving it directly? If yes, why?

Thanks a lot in advance![/ltr]

Bunuel

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

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

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

Complete the square for $x^2 - 12 x - 64$:

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

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

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

Show Spoiler

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GMAT-Club-Forum-v9g7ur1j.png [ 16.16 KiB | Viewed 2769 times ]

Bunuel

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Sujithz001

Hey Bunuel / KarishmaB / gmatophobia

Might be a beginner question - I'm facing difficulty understanding the solution from this step,

[ltr]
1. Why are we manipulating the quadratic equation, rather the solving it upfront?

2. Would we not reach the soln. by solving it directly? If yes, why?

Thanks a lot in advance![/ltr]

Bunuel

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

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

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

Complete the square for $x^2 - 12 x - 64$:

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

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

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

To solve, we need an equation, like $-40(x^2 - 12x - 64) = 0$, which could be solved for x to get specific values: x = -4 or x = 16. However, in this case, we are dealing with an expression, $-40(x^2 - 12x - 64)$, for which we need to find the maximum value.

To do this, we manipulated the expression to break it into a sum of two parts: a non-positive term $-40(x - 6)^2$ and 4,000. This setup helps us see that the maximum value of the expression occurs when $-40(x - 6)^2$ is 0, meaning the overall maximum is 0 + 4,000 = 4,000.

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Globethrotter

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Bunuel A faster way to get to the answer is with a little bit different approach without actually solving any equation.
Just write down the possible pairs as shown below and observe the point at which the pattern starts to reverse. That is the inflection point and thus the point of maximum value.

See below, the green value is the inflection point. Observe the values start inversing at that point(for eg. 22*180 becomes 18*220)

32	80
30	100
28	120
26	140
24	160
22	180
20	200
18	220
16	240
14	260
12	280
10	300

So the max value is 20*200 = 4000, Store currently selling at 32*80 = 2560.
Max amount by which store can increase its revenue = 4000 - 2560 = 1440.

Answer is A