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Manufacturing an automobile requires painting, drying, and polishing. Epsilon Motor Company produces three types of cars: the Delta, the Beta, and the Sigma.

Each Delta requires 7 hours for painting, 2 hours for drying, and 5 hours for polishing.
Each Beta requires 16 hours for painting, 5 hours for drying, and 6 hours for polishing.
Each Sigma requires 6 hours for painting, 5 hours for drying, and 3 hours for polishing.

If the company has 165 hours for painting, 94 hours for drying, and 89 hours for polishing per month, how many of each type of car are produced?

Select the correct choice below and fill in any answer boxes within your choice.

A. The Epsilon Motor Company produces [tex] \square [/tex] Deltas, [tex] \square [/tex] Betas, and [tex] \square [/tex] Sigmas in a month.
(Simplify your answers.)

B. There are infinitely many combinations of each type of car that can be produced. Using ordered triplets, the solution can be written as [tex] \{(D, B, S) \mid D= \square, B= \square S, \text{ S any real number} \} [/tex].
(Simplify your answers. Type expressions using S as the variable as needed.)

C. There is no possible combination of each type of car that can be produced that meets the criteria.

Sagot :

GinnyAnswer
To determine the number of Delta (D), Beta (B), and Sigma (S) cars that can be produced in a month, we can set up and solve the system of linear equations based on the information provided about the hours required for painting, drying, and polishing.

We have the following constraints:
1. Painting: [tex]\(7D + 16B + 6S = 165\)[/tex]
2. Drying: [tex]\(2D + 5B + 5S = 94\)[/tex]
3. Polishing: [tex]\(5D + 6B + 3S = 89\)[/tex]

To solve this system of equations, we will use the method of elimination or substitution.

Let's use the elimination method:

1. Multiply the second equation by 2.5 (to match the drying term with the polishing equation):
[tex]\[ 2.5(2D + 5B + 5S) = 2.5 \times 94 \][/tex]
[tex]\[ 5D + 12.5B + 12.5S = 235 \quad \text{(Eq 4)} \][/tex]

2. Now, subtract the polishing equation (Eq 3) from this new equation (Eq 4):
[tex]\[ (5D + 12.5B + 12.5S) - (5D + 6B + 3S) = 235 - 89 \][/tex]
[tex]\[ 6.5B + 9.5S = 146 \quad \text{(Eq 5)} \][/tex]
[tex]\[ 13B + 19S = 292 \quad \text{(simplified by multiplying both sides by 2)} \][/tex]

3. Now, we can take the first equation and the new equation from step 2:
[tex]\[ 7D + 16B + 6S = 165 \quad \text{(Eq 1)} \][/tex]
[tex]\[ 13B + 19S = 292 \quad \text{(Eq 5)} \][/tex]

4. Let's express [tex]\(B\)[/tex] in terms of [tex]\(S\)[/tex] from Eq 5:
[tex]\[ 13B = 292 - 19S \][/tex]
[tex]\[ B = \frac{292 - 19S}{13} \][/tex]

5. Substitute this [tex]\(B\)[/tex] back into Eq 1 to find [tex]\(D\)[/tex] in terms of [tex]\(S\)[/tex]:
[tex]\[ 7D + 16\left(\frac{292 - 19S}{13}\right) + 6S = 165 \][/tex]
[tex]\[ 7D + \frac{16 \times 292 - 16 \times 19S}{13} + 6S = 165 \][/tex]
[tex]\[ 7D + \frac{4672 - 304S}{13} + 6S = 165 \][/tex]
[tex]\[ 7D + \frac{4672 - 304S + 78S}{13} = 165 \][/tex] (Combining like terms)
[tex]\[ 7D + \frac{4672 - 226S}{13} = 165 \][/tex]
[tex]\[ 7D + \frac{4672 - 226S}{13} = 165 \][/tex]
[tex]\[ 7D \cdot 13 + 4672 - 226S = 165 \cdot 13 \][/tex]
[tex]\[ 91D + 4672 - 226S = 2145 \][/tex]
[tex]\[ 91D + 4672 - 226S = 2145 \][/tex]
[tex]\[ 91D = 2145 - 4672 + 226S \][/tex]
[tex]\[ 91D = -2527 + 226S \][/tex]
[tex]\[ D = -\frac{2527}{91} + \frac{226}{91} S \][/tex]
[tex]\[ D = -27.8 + 2.48S \][/tex]

Thus, we can express [tex]\(D\)[/tex] and [tex]\(B\)[/tex] in terms of [tex]\(S\)[/tex] as:
[tex]\[ D = -27.8 + 2.48S \][/tex]
[tex]\[ B = 22.4615 - 1.46S \][/tex]

6. These results indicate relationships dependent on [tex]\(S\)[/tex], which means that for any value of [tex]\(S\)[/tex], there will be corresponding values of [tex]\(D\)[/tex] and [tex]\(B\)[/tex] that satisfy all three equations. This implies that there are infinitely many solutions forming an infinite set of ordered pairs (D, B, S).

Hence, the correct answer is:

B. There are infinitely many combinations of each type of car that can be produced. Using ordered triplets, the solution can be written as [tex]\( \{(D, B, S) | D=-27.8 + 2.48S, B=22.4615 - 1.46S, S \text{ any real number}\} \)[/tex]
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