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Two models of a product - Regular (X) and Deluxe (Y) - are produced by a company. A linear programming model is used to determine the production schedule. The formulation is as follows:


Maximize profit = 50X + 60Y


Subject to:


8X + 10Y=800 (labor hours)


X + Y=120 (total units demanded)


4X + 5Y =500 (raw materials)


all variables =0


The optimal solution is X = 100, Y = 0


How many units of the regular model would be produced based on this solution?


A. 0

B. 50

C. 100

D. 120

Respuesta :

Answer:

The correct option is;

C. 100

Step-by-step explanation:

Here we have

Maximum profit = 50X + 60Y

8X + 10Y ≤ 800 (labor hours)

X + Y ≤ 120 (total units demanded)

4X + 5Y ≤ 500 (raw materials)

All variables ≥ 0

Then we have

(i) 8X + 10Y ≤ 800

X = 0 or 100

Y = 80 or 0

(ii) X + Y ≤ 120

X = 0 or 120

Y = 120 or 0

(iii) 4X + 5Y ≤ 500

X = 0 or  125

Y = 100 or 0

from a chart of the above values we have in the most possible region, we have

Y intercept at X = 0 and Y = 80

Origin at  X = 0, Y = 0

X intercept at X = 100 and Y = 0

Therefore, to maximize profit we have z(50X + 60Y)

At Y intercept (0, 80) = 4800

At origin = 0

At X intercept (100, 0) = 5000

Therefore, the number of units of regular model, X for maximum profit should be 100.

lhmarianateixeira

In this exercise we have to analyze the functions and mark the alternative that best corresponds, that way we have:

The correct option is C

Here we have maximum profit equal to  [tex]50X + 60Y[/tex]

[tex]8X + 10Y \leq 800 \\X + Y \leq 120 \\4X + 5Y \leq 500[/tex]

Then we have:

(i) The first equation is:

[tex]8X + 10Y \leq 800\\X = 0 \ or \ 100\\Y = 80 \ or \ 0[/tex]

(ii) The second equation is:

[tex]X + Y \leq 120\\X = 0 \ or\ 120\\Y = 120 \ or \ 0[/tex]

(iii) The third equation is:

[tex]4X + 5Y \leq 500\\X = 0 \ or \ 125\\Y = 100 \ or \ 0[/tex]

From a chart of the above values we have in the most possible region, we have:

  • Y intercept at X = 0 and Y = 80
  • Origin at  X = 0, Y = 0
  • X intercept at X = 100 and Y = 0

Therefore, to maximize profit we have:

  • At Y intercept [tex](0, 80) = 4800[/tex]
  • At origin = 0
  • At X intercept [tex](100, 0) = 5000[/tex]

Therefore, the number of units of regular model, X for maximum profit should be 100.

See more about functions at brainly.com/question/5245372

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