A major manufacturer of MP3 players wants to produce both black and wh

Question

A major manufacturer of MP3 players wants to produce both black and white plastic cases for its new 32 GB MP3 line. The manufacturer wants to spend a maximum of $200,000 dollars on case production. White plastic cases cost $2 more per case to produce than black plastic cases.

Using W to represent the cost of producing each white case and B to represent the cost of producing each black case, select the expression that reflects the cost of producing one white case, and select the expression that reflects the maximum number of each type of case that can be produced within the company’s $200,000 budget if the same number of black and white MP3 cases are made. Make only two selections, one in each column.

poojaarora1818 · Answer

Solution:

Given in the question stem:

1. The manufacturer can spend $200,000 on total case production.
2. The cost of producing each white case = W
3. The cost of producing each black case = B
4. White plastic cases cost $2 more per case to produce than black plastic cases. W = B + 2

By using the formula Average =  \frac{Total number of observations}{No of observations} , we can solve the question.

1.  The   cost of producing one white case = W = B + 2.

2. The maximum number of each type of case that can be produced within the company’s $200,000 budget if the same number of black and white MP3 cases are made = B + 2 + B = 200,000 =  \frac{200,000}{2B + 2}

gabgol1998 · Answer

Let's plug in numbers.

8 for each Black (B)

Cost of one white case.

8 + 2 = 10

Option 1 -

B + 2. (Correct)

Second part:

Let's say the manufacturer would produce 20 in total, then 10 black and 10 white.

Apply the formula Cost x Quantitity = Total Cost considering the two products.

8x10 + 10x10 = 180 (too small, try 22 in total, 11 each case)

8x11 + 10x11 - 198 (we can't go further, otherwise it would be higher than 200.)

Checking in the answers:

Option 2 -

2W + B

2x10 + 8 = 28 (not eleven)

Option 3 -

200.000/(2B + 2)

200.000/18 = 11111.11 (Correct)

Since we reduced three 0's in the original number, this answer is actually giving us 11.11 (consider 200 in the numerador, the only difference is the 0's)
Moreover, we can't have 11.11 as maximum, we only can produce 11 or 12, hence we have 11 as answer here.

Cost of Producing one white case	Maximum Number of each colour case	Expression--------------
		\(B+2\)
		\(2W+B\)
		\(\frac{200,000}{2B+2}\)
		\(200,000(W+B)\)
		\(\frac{200,000B}{2W}\)

A major manufacturer of MP3 players wants to produce both black and wh

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A major manufacturer of MP3 players wants to produce both black and wh

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