M40-98

Question

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

Show Answer

Official Answer

A

A. 1,440
B. 1,600
C. 2,460
D. 3,920
E. 4,000

Bunuel · Answer

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

Show Answer

Official Answer

A

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

=-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

shyam0701 · Answer

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!

Mantrix · Answer

Two methods By which we can solve this question
 
 1. Basic Method:

just take the cases out
1. Price: 32$, Shoes: 80 this gives the revenue as 2560$
2. Price: 30$, Shoes: 100 this gives the revenue as 3000$
3. Price: 28$, Shoes: 120 this gives the revenue as 3360$
4. Price: 26$, Shoes: 140 this gives the revenue as 3640$
5. Price: 24$, Shoes: 160 this gives the revenue as 3840$
6. Price: 22$, Shoes: 180 this gives the revenue as 3960$
7. Price: 20$, Shoes: 200 this gives the revenue as 4000$
8. Price: 18$, Shoes: 220 this gives the revenue as 3960$(the amount starts to reduce)

so max revenue we can get from the 7th case i.e 4000$, 
By what amount it increase: (4000-2560)$ that comes out to be 1440$

2. 2nd Method

By using maxima/Minima Concept

We can form the equation as Revenue = (80+20x) * (32-2x) [X is the number of iteration)
for max min, differentiation of the equation must be zero
hence, 
Revenue = -40x^2-160x+640x+2560 = revenue
hence, 
dr/dx = -80x+480 = 0
so x=6

so Revenue by putting x in the equation comes out be: 4000$
By what amount it increase: (4000-2560)$ that comes out to be 1440$

Spectre26 · Answer

Yep did the same thing !!
In few questions d/dx to find maxima point helps.

pyreddy · Answer

I did not quite understand the solution. Brunel can you please explain. Here 2 dollars for each pair so the new price will become 30 . 
So new revenue100*30 =3000 
Old revenues 80*32 = 2560

Bunuel · Answer

You're right, when the price drops from $32 to $30 (i.e., x = 1), the store sells 100 pairs, and:

New revenue = 100 * 30 = 3000 
 Old revenue = 80 * 32 = 2560 
 Increase = 440

But the question asks for the  maximum  possible increase in revenue, not just at x = 1. The official solution checks all possible values of x and finds that the maximum happens when x = 6.

Here's what happens at x = 6:

New price = 32 - 2*6 = 20 
 New sales = 80 + 20*6 = 200 
 New revenue = 20 * 200 = 4000 
 Old revenue = 32 * 80 = 2560 
 Maximum increase = 4000 - 2560 = 1440

That’s why the final answer is A. 1,440.

Manvi01 · Answer

I like the solution - it’s helpful.

anushree01 · Answer

Is maxima & minima by diffrentiation- a part of gmat syllabus ?  Bunuel  if you could pls tell!
Thanks!

Bunuel · Answer

It wouldn’t hurt if you know how to differentiate. However, GMAT algebra questions asking for a maximum or minimum can always be solved without differentiation, as shown in the solution above.

BinodBhai · Answer

This is a great question that’s helpful for learning.

ADhanjal · Answer

Hi  Bunuel ,

I did not understand why you used 'completing the squares' method, do I have to use this method every time a question asks us to find the maximum/minimum values?

Bunuel · Answer

No, you don’t always have to use completing the square. That’s just one standard way to find the maximum or minimum of a quadratic. Here the revenue function (32 - 2x)(80 + 20x) is quadratic in x. Any quadratic has a parabola shape, so its maximum or minimum is at the vertex. You can find that vertex either by completing the square, or if you know that by using -b/2a from the standard ax^2 + bx + c form. Both give the same result. You can check alternative solution here: https://gmatclub.com/forum/a-shoe-store ... 98695.html

harshitasinghal · Answer

2. 2nd Method

By using maxima/Minima Concept

We can form the equation as Revenue = (80+20x) * (32-2x)  
 for max min, differentiation of the equation must be zero 
 hence, 
 Revenue = -40x^2-160x+640x+2560 = revenue 
 hence, 
 dr/dx = -80x+480 = 0 
 so x=6

so Revenue by putting x in the equation comes out be: 4000$ 
 By what amount it increase: (4000-2560)$ that comes out to be 1440$

Can you please explain the dr/dx = -80x+480 = 0. I didn't understand this.

Bunuel · Answer

That solution uses derivatives, which you don’t need for the GMAT. If you’re not comfortable with calculus, it’s better to stick with the standard non-calculus method shown above. Also, you can check some other methods here: https://gmatclub.com/forum/a-shoe-store ... 98695.html

Dream009 · Answer

My way of thinking 32*80 = 2400+160 = 2560
Now (32-2x)(80+20x) = revenue = to maximize differentiate

32(20) - 2*80 - 40*2x = 0
divide by 20 
32 - 8 = 4x
24 = 4x
x = 6
Now new revenue is 20*200 = 4000 
So 4000 - 2560 ~ 1500 Ans B

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