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

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  

Then we have:

(i) The first equation is:

(ii) The second equation is:

(iii) The third equation is:

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

Therefore, to maximize profit we have:

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

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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.

i would pic false cause it goes down towards negative

step-by-step explanation: