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Studentka2010 [4]
3 years ago
7

Can one fifth be simplified

Mathematics
1 answer:
Svetradugi [14.3K]3 years ago
3 0
No, and never, cus there ain't any numbers that are greater than 1 that can go into both of them.
<span>so basically if the numerator is 1, you can't simplify it anymore. Is Impossible.</span><span />
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A manufacturing company produces 3 different products A, B, and C. Three types of components, i.e., X, Y, and Z, are used in the
Murljashka [212]

Answer:

Step-by-step explanation:

Using the Excel Formula:

Decision    Variable        Constraint              Constraint

A                     65                          65                         100

B                     80                          80                         80

C                     90                         90                          90

                      14100                    300                        300

= (150 *B3)+(80*B4) +(65*B5)-(100-B3+80-B4+90-B5)*90

Now, we have:

Suppose A, B, C represent the number of units for production A, B, C which is being manufactured

                             A              B                  C                Unit price

Need of X          2                 1                   1                     $20

Need of Y           2                3                  2                    $30

Need of Z           2                2                  3                    $25

Price of  

manufac -      $200          $240            $220      

turing

Now,  for manufacturing one unit of A, we require 2 units of X, 2 units of Y, 2 units of Z are required.  

Thus, the cost or unit of manufacturing of A is:

$20 (2) + $30(2) + $25(2)

$(40 + 60 + 50)

= $150

Also, the market price of A = $200

So, profit = $200 - $150 = $50/ unit of A

Again;

For manufacturing one unit of B, we require 1 unit of X, 3 units of Y, and 2 units of Z are needed and they are purchased at $20, $30, and 425 each.

So, total cost of manufacturing a unit of B is:

= $20(1) + $30(3) + $25(2)

= $(20 + 90+50)

= $160

And the market price of B = $240

Thus, profit = $240- $160  

profit = $80

For manufacturing one unit of C, we have to use 1 unit of X, 2 unit of Y, 3 units of Z are required:

SO, the total cost of manufacturing a unit of C is:

= $20 (1) + $30(2) + $25(3)

= $20 + $60 + $25

= $155

This, the profit = $220 - $155 = $65

However; In manufacturing A units of product A, B unit of product B & C units of product C.

Profit  --> 50A + 80B + 65C

This should be provided there is no penalty for under supply of there is under supply penalty for A, B, C is $40

The current demand is:

100 - A

80 - B

90 - C respectively

So, the total penalty

{(100 - A) + (80 - B) +(90 - C) } + \$40

This should be subtracted from profit.

So, we have to maximize the profit  

Z = 50A + 80B + 65C = {(100 -A) + (80 - B) + (90 - C)};

Subject to constraints;

we have the total units of X purchased can only be less than or equal to 300 due to supplies capacity

Then;

2A + B +C \le 300 due to 2A, B, C units of X are used in manufacturing A, B, C units of products A, B, C respectively.

Next; demand for A, B, C will not exceed 100, 80, 90 units.

Hence;

A \le 100

B \le 80

C \le 90

 

and A, B, C \ge 0 because they are positive quantities

The objective is:

\mathbf{Z = 50A + 80B + 65 C - (100 - A + 80 - B + 90 - C) * 40}

A, B, C \to Decision Varaibles;

Constraint are:

A \le 100 \\ \\  B \le 100 \\ \\ C \le 90 \\ \\2A + B + C \le 300 \\ \\ A,B,C \ge 0

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