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A factory can produce two products, x and y, with a profit approximated by P=14x+22y-900. The production of y must exceed the production of x by at least 100 units. Moreover, production levels are limited by the formula x+2y<1400.

a) Identify the vertices of the feasible region.
b) What production levels yield the maximum profit, and what is the maximum profit?

Respuesta :

wegnerkolmp2741o

Answer:

a)  The vertices of the feasible region are  (0,100) (0,700) (400,500)

The minimum profit is at (0,100)  and the maximum profit is at (400,500)

Step-by-step explanation:

P=14x+22y-900  where p is profit

y > x +100    y must exceed the production of x by at least 100 units

x+2y<1400

x>0

y>0

We cannot produce negative quantities

Substitute y = x+100 into x+2y <1400

x+2(x+100) < 1400

x+2x+200 <1400

3x+200<1400

Subtract 200 from each side

3x<1200

Divide by 3

x<400

y = x+100

y = 400+100

y = 500

(400,500)

y > x +100

when x=0  y > 100  

x+2y <1400

0+2y <1400

 2y <1400

y <700

When x=0  y = 700

a)  The vertices of the feasible region are  (0,100) (0,700) (400,500)

b)  Maximum and minimum profit occur at the vertices.

P=14x+22y-900

P(0,100) = 14(0) +22(100)-900 =2200-900=1300

P(0,700) = 14(0) +22(700)-900 =15400-900=14500

P(400,500) = 14(400) +22(500)-900 =5600+11000-900=15700

The minimum profit is at (0,100)  and the maximum profit is at (400,500)

Ver imagen wegnerkolmp2741o
MrRoyal

When profits are maximized, it means we want to get the highest possible profit from a function.

  • The vertices of the feasible region are: [tex]\mathbf{(x,y) = \{(0,100), (0,700),(400,500)\}}[/tex]
  • The production level that yields maximum profit is [tex]x = 400; y = 500[/tex]
  • The maximum profit is #17100

Production of y must exceed x by 100 means

[tex]y > x + 100[/tex]

The production level is given as:

[tex]x + 2y < 1400[/tex]

Since x and y represent products, then:

[tex]x, y > 0[/tex]

So, we have:

[tex]P = 14x + 22y - 900[/tex] --- the objective function

Subject to

[tex]y > x + 100[/tex]

[tex]x + 2y < 1400[/tex]

[tex]x, y > 0[/tex]

See attachment for the graph of the subjects

(a) The vertices of the feasible region

From the attached image, the vertices of the feasible region are:

[tex]\mathbf{(x,y) = \{(0,100), (0,700),(400,500)\}}[/tex]

(b) The maximum profit

Recall that: [tex]x, y > 0[/tex]

This means that, we can only make use of:

[tex]\mathbf{(x,y) = \{(400,500)\}}[/tex]

So;

The production levels that yield maximum profit are:

[tex]x = 400; y = 500[/tex]

Substitute these values in [tex]P = 14x + 22y - 900[/tex], to calculate the maximum profit

[tex]P = 14x + 22y - 900[/tex]

[tex]P = 14 \times 500 + 22 \times 500 - 900[/tex]

[tex]P = 17100[/tex]

Hence, the maximum profit is: #17100

Read more about maximizing profits at:

https://brainly.com/question/11206462

Ver imagen MrRoyal

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