A company can use two workers to manufacture product 1 and product 2 during a business slowdown. Worker 1 will be available for 20 hours and worker 2 for 24 hours. Product 1 will require 5 hours of labor from worker 1 and 3 hours of specialized skill from worker 2. Product 2 will require 4 hours from worker 1 and 6 hours from worker 2. The finished products will contribute a net profit of $60 for product 1 and $50 for product 2. At least two units of product 2 must be manufactured to satisfy a contract requirement. Formulate a linear program to determine the profit maximizing course of action. (Hint: the simplest formulation assigns one decision variable to account for the number of units of product 1 to produce and the other decision variable to account for the number of units of product 2 to produce.)

Answers

Answer 1

Answer:

Answer:

The linear problem is to maximize $Z = C_ {1} X_ {1} + C_ {2}X_ {2} = 60X_ {1} + 50X_ {2}$ , s.a.

subject to

$\frac {1} {5} X_ {1} + \frac {1} {4} X_ {2} \leq 20\n\n\frac {1} {3} X_ {1} + \frac {1} {6} X_ {2} \leq 24\n\nX_ {2} \geq 2\n\nX_ {1}, X_ {2} \geq 0$

Step-by-step explanation:

Let the decision variables be:

$X_ {1}$ : number of units of product 1 to produce.

$X_ {2}$ : number of units of product 2 to produce.

Let the contributions be:

$C_ {1} = 60\n\nC_ {2} = 50$

The objective function is:

$Z = C_(1) X_(1)+ C_(2)X_(2) = 60X_ {1} + 50X_ {2}$

The restrictions are:

$\frac {1} {5} X_ {1} + \frac {1} {4} X_ {2} \leq 20\n\n\frac {1} {3} X_ {1} + \frac {1} {6} X_ {2} \leq 24\n\nX_ {2} \geq 2\n\nX_ {1}, X{2} \geq 2\n\n$

The linear problem is to maximize $Z = C_ {1} X_ {1} + C_ {2}X_ {2} = 60X_ {1} + 50X_ {2}$ , s.a.

subject to

$\frac {1} {5} X_ {1} + \frac {1} {4} X_ {2} \leq 20\n\n\frac {1} {3} X_ {1} + \frac {1} {6} X_ {2} \leq 24\n\nX_ {2} \geq 2\n\nX_ {1}, X_ {2} \geq 0$

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Need help ASAP if y’all could pls help me out that would be great

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hi! here’s what i’d put for the sentences to fill in the blanks:

- 1 ten is equal to 10 ones. 1 ten is 10 times the value of 1 one.

- 1 hundred is equal to 10 tens. 1 hundred is 10 times the value of 1 ten.

- 1 thousand is equal to 10 hundreds. 1 thousand is 10 times the value of 1 hundred.

- 1 ten-thousand is equal to 10 thousands. 1 ten-thousand is 10 times the value of 1 thousand.

- 1 hundred-thousand is equal to 10 ten-thousands. 1 hundred-thousand is 10 times the value of 1 ten-thousand.

Place Value Chart:

blank space before 3 is 2

hope this helped!:)

7th grade math help me plzzzz

Answers

Answer: y = 3x

Step-by-step explanation: Proportional means that the y-value will always be three times the x-value--

Or whatever the coefficient of x is in this type of equation.

Monty can use the number line to find an equivalent fraction with a denominator greater than 6

Answers

Yes, Monty can use the number line to find an equivalent fraction with a denominator greater than 6.

For Example,

Consider $(5)/(7)$

The equivalent fraction of $(5)/(7)$ is $(25)/(35)$ .

So, yes you can represent $(5)/(7)$ on a number line by putting 6 lines between 0 and 1 and Can Represent $(25)/(35)$ by putting 34 lines between 0 and 1.

There is no effect of denominator to find equivalent fraction of any rational number, whether the denominator is greater than 6 or less than 6, but denominator should not be equal to Zero.

We can find equivalent fraction of any rational number , the denominator of that rational number should not be equal to Zero.

Is this math? What's the question ?

(PLEASE HELP!!!)Which shows the list of numbers in order from least to greatest?