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Question 1 A small company manufactures three different electronic components for computers. Component A requires two hours of fabrication and 1 hour of assembly. Component B requires 3 hours of fabrication and 1 hour of assembly. Component C requires 2 hours of fabrication and 2 hours of assembly. The company has up to 1,000 labor-hours of fabrication time and 800 labor-hours of assembly time available per week. Due to storage restrictions, the combined total number of components produced each week cannot exceed 420. The profit on each component A, B, and C is $7, $8 and $10 respectively. How many components of each type should they company manufacture each week in order to maximize profit, assuming that all components can be sold

Sagot :

Answer: A should be A=200, B should be B=0, and component C should be C=350. This will lead to a maximum profit of$4550.

Explanation:

Let A, B, and C be the number of components of types A, B, and C that the company should manufacture each week. The aim is to maximize the profit. Profit per unit of A, B, and C is $7, $8, and $10, respectively. So, formulate the objective function as follows :MaxP=7A+8B+10C

Lanuel

Based on the calculations, the maximum profit for this small company is equal to $4,400.

How to calculate the maximum profit​?

  • Let the fabrication time be x.
  • Let the assembly be y.

Next, we would represent the given parameters by using a table as follows:

          x          y       profit    

A        2          1           7

B        3          1           8

C        2          2         10

        1000    800

From the table above, we subject to have the following system of equations:

2A + 3B + 2C = 1000

A + B + 2C = 800

Mathematically, the maximum profit for this company is given by:

P = 7A + 8B + 10C

By using matrix, the maximum profit and number of electronic components produced per week would be calculated as follows:

[tex]\left[\begin{array}{cccccccc}2&3&2&1&0&0&1000\\1&1&2&0&1&0&800\\-7&-8&-10&0&0&1&0\end{array}\right] \\\\\\\left[\begin{array}{cccccccc}1&2&0&1&-1&0&200\\\frac{1}{2} &\frac{1}{2}&1&0&\frac{1}{2}&0&400\\-2&-3&0&0&5&1&4000\end{array}\right]\\\\\\\left[\begin{array}{cccccccc}\frac{1}{2}&1&0&\frac{1}{2}&-\frac{1}{2}&0&100\\\frac{1}{4} &0&1&\frac{-1}{4}&\frac{3}{4}&0&350\\\frac{-1}{2}&0&0&\frac{3}{2}&\frac{7}{2}&1&4300\end{array}\right][/tex]

[tex]\left[\begin{array}{cccccccc}1&2&0&-1&-1&0&200\\0 &\frac{-1}{2}&1&\frac{-1}{2}&1&0&300\\0&1&0&2&3&1&4400\end{array}\right][/tex]

Thus, we have:

A = 200 components.

B = 0 components.

C = 300 components.

Maximum profit = $4,400.

Read more on maximum profit here: https://brainly.com/question/22419316

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