men's blazer

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gmatrix: Master | Next Rank: 500 Posts; Posts: 108; Joined: Fri Jul 09, 2010 8:15 am; Location: 127.0.0.1; Thanked: 15 times

by gmatrix » Tue Oct 26, 2010 1:47 am

A certain clothing manufacturer makes only two types of men's blazer: cashmere and mohair. Each cashmere blazer requires 4 hours of cutting and 6 hours of sewing. Each mohair blazer requires 4 hours of cutting and 2 hours of sewing. The profit on each cashmere blazer is $40 and the profit on each mohair blazer is $35. How many of each type of blazer should the manufacturer produce each week in order to maximize its potential weekly profit on blazers?

1) The company can afford a maximum of 200 hours of cutting per week and 200 hours of sewing per week.

2) The wholesale price of cashmere cloth is twice that of mohair cloth.

OA:later

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Source: Beat The GMAT — Data Sufficiency | Tue Oct 26, 2010 1:47 am

kmittal82: Master | Next Rank: 500 Posts; Posts: 324; Joined: Mon Jul 05, 2010 6:44 amFollowed by: 3 members; Location: London; Thanked: 70 times

by kmittal82 » Tue Oct 26, 2010 2:29 am

Wow, what a question!

Right, lets eliminate 2 to start with, nothing is mentioned about the price of cloth.

So, lets look at 1

1) Max cutting hours = 200 , max seweing hours = 200

a) If the retailer maximizes cashmere production, then max number of cashmere blazers he can make = 33 (200/6 = 33.33)
this means 33x4 = 138 hours cutting time and 198 hours sewting time is taken, which leaves just enough time to make 1 mohair blazer

Thus, profit = 33 cashmere + 1 mohair 33x40 + 35 = 1355

b)

If the retailer maximizes mohair blazers, then he can make a max of 50 mohair blazers (200/4 = 50). This takes up all cutting time, although leaving sewing time, but its not enough to make any cahsmeres

Thus, profit = 50 x 35 = 1750

c) If the retailer decides to make both.

Now, total time to make 1 mohair and 1 cashmere suit = 8 hours cutting + 8 hours sewing

total time available = 200hours cutting + 200 hours sewings

Thus, the retailer can evenly split this and make 25 mohairs and 25 cashmeres

Profit = 25x40 + 25x35 = 1875

Out of the 3 possibilities, last one maximises the profits, which shows us maximum profits will be attained only when the retailers make both types of suits.

This is where I get lost, we have 2 variables (i.e. number of suits of each type), and we need to find a combination which maximises the profits, but only 1 equation.

Shot in the dark, but I think its (E)

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kvcpk: Legendary Member; Posts: 1893; Joined: Sun May 30, 2010 11:48 pmFollowed by: 7 members; Thanked: 215 times

by kvcpk » Tue Oct 26, 2010 2:35 am

Hi Mittal,

You were right .. till the end, Except for your answer. IMO A.

The maximum profit is when he makes 25 suits of each kind.

We need to make the complete use of 200 hours. So,

4c + 4m = 200
6c + 2m = 200

Let us equate both of them:
4c + 4m = 6c + 2m
2c = 2m
c=m

By solving we get, c=m=25.

hence pick A.

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don't be afraid of failure and don't abandon it.
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kmittal82: Master | Next Rank: 500 Posts; Posts: 324; Joined: Mon Jul 05, 2010 6:44 amFollowed by: 3 members; Location: London; Thanked: 70 times

by kmittal82 » Tue Oct 26, 2010 2:45 am

kvcpk wrote:Hi Mittal,

You were right .. till the end, Except for your answer. IMO A.

The maximum profit is when he makes 25 suits of each kind.

We need to make the complete use of 200 hours. So,

4c + 4m = 200
6c + 2m = 200

Let us equate both of them:
4c + 4m = 6c + 2m
2c = 2m
c=m

By solving we get, c=m=25.

hence pick A.

Ofcourse, max utilization of time would reap the profits!

Thanks for your explanation

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vijchid: Junior | Next Rank: 30 Posts; Posts: 17; Joined: Wed Sep 29, 2010 8:14 am; Thanked: 1 times

by vijchid » Tue Oct 26, 2010 1:07 pm

This seems to be a problem right out of a linear programming text book.
Maximize 40C + 35M
withe constraints
4C+4M<=200
6C+2M<=200
Complete utilization of time might not be the solution for these kinds of problem. If you draw the two constraints on a graph the feasible solution region is a simple polygon and hence a global optimum can be found... This problem seems to be out of scope for GMAT? Where did you find this one?

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diebeatsthegmat: Legendary Member; Posts: 1119; Joined: Fri May 07, 2010 8:50 amFollowed by: 3 members; Thanked: 29 times

by diebeatsthegmat » Wed Oct 27, 2010 9:59 am

gmatrix wrote:A certain clothing manufacturer makes only two types of men's blazer: cashmere and mohair. Each cashmere blazer requires 4 hours of cutting and 6 hours of sewing. Each mohair blazer requires 4 hours of cutting and 2 hours of sewing. The profit on each cashmere blazer is $40 and the profit on each mohair blazer is $35. How many of each type of blazer should the manufacturer produce each week in order to maximize its potential weekly profit on blazers?

1) The company can afford a maximum of 200 hours of cutting per week and 200 hours of sewing per week.

2) The wholesale price of cashmere cloth is twice that of mohair cloth.

OA:later

a+
what is the answer and its explaination too cos my method to get it is so long and time consuming

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

by Rahul@gurome » Wed Oct 27, 2010 9:23 pm

Let the number of cashmere blazers manufactured per week be c and the number of mohair blazers manufactured per week be m.
Hence profit on all cashmere blazers is $40c and profit on all mohair blazers is $35m
Also the time spent per week on cutting is 4c + 4m and the time spent per week on sewing is 6c+2m.
We need to maximize 40c+35m.
Let us first consider statement (1) alone.
It says 4c+4m<=200 and 6c+2m<=200.
So c+m <= 50 and 3c+m<= 100.
Or 13c+13m <= 650 and 3c+m <= 100.
Or adding the inequalities we get that 16c+14m<=750.
Or 8c+7m<=375. This is same as 40c+35m<=1875.
Note that c and m are positive integers.
Since we need the maximum value of 40c+35m which is 1875, we look for integral values of c and m which satisfy the equation 40c+35m = 1875 or 8c+7m = 375.
Note that m+c<=50. This means 8m+8c<=400.
Or 375-7m+8m<=400.
Or m<=25.
So start testing with m = 25 in the equation 8c+7m = 375
We see that m = 25 and c = 25 satisfy this equation.
So (1) alone is sufficient.
Now (2) alone is obviously not sufficient since it is not giving any information on time spent on the blazers.

The correct answer is hence A.

Rahul Lakhani
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