Question

A company produces mopeds and bicycles. It must produce at least 10 mopeds per month. The company has the equipment to produce only 60 mopeds. It also can produce only 120 bicycles. The production of mopeds and bicycles cannot exceed 160. The profit on a moped is $134 and on a bicycle $20. How many of each should be manufactured per month to maximize profit?

(Use x = mopeds; y = bicycles)


3. State the Objective Function in the linear programming problem given:

A. P = 20x + 134y

B. P = 134x + 20y

C. P = 60x + 120y

D. P = 120x + 60y

4. Determine the vertices of the feasible region, given the constraints. (Use Desmos to help with graphing!)

Constraints (for this linear programming problem):
A. (0, 0) F. (160,0)
B. (10,120) G. (60,100)
C. (60, 0) H. (0,120)
D. (60,120) I. (40,120)
E. (10, 0) J. (10,150)

5. How many of each product should be manufactured per month to maximize profit?
A. 60 mopeds, 120 bicycles
B. 160 mopeds, 0 bicycles
C. 60 mopeds, 100 bicycles
D. 40 mopeds, 120 bicycles

PLEASE SHOW WORK.
WILL GIVE BRAINLIEST!
THE PICTURE IS FOR QUESTION 4

Answer 1

Answer:

3. B. P = 134x + 20y

4. J. (10, 150), I(40, 120), G(60, 100), C(60, 0) and E(10, 0)

5. C. 60 mopeds 100 bicycles

Step-by-step explanation:

3. The given information are;

The number of mopeds the company must produce per month = 10

The number of mopeds the company can produce month = 60

The number of bicycles the company can produce per month = 120

The total sum of the produced mopeds and bicycles ≤ 160

The profit on a moped = $134

The profit on a bicycle = $20

Given that the number of moped produced = x and the number of bicycles produced = y

Therefore, we have;

Profit, P = $134 × x + $20 × y

Which gives;

B. P = 134x + 20y

4) From Desmos, using the following constraints;

x + y ≤ 160

10 ≤ x ≤ 60

0 ≤ y ≤ 120 we have;

The vertices of the constraint for the feasible region are;

J. (10, 150), I(40, 120), G(60, 100), C(60, 0) and E(10, 0)

5) We note that since the profit from each moped is more than the profit from the sale of each bicycle, the maximum possible number of moped should be produced, while the rest should be used to produce bicycles

Therefore, given that from the vertices, the maximum possible number of moped = 60, the number of bicycles to be produced should be 160 - 60 = 100 bicycles

Which gives;

C. 60 mopeds 100 bicycles.

Answer 2

3. The objective function should be option B. P = 134x + 20y

4. The vertices of the feasible region, given the constraints should be J(10, 150), I(40, 120), G(60, 100), C(60, 0) and E(10, 0)

5. The number of each product that should be manufactured per month should be option C. 60 mopeds 100 bicycles

Objective function, vertices, and number of each product:

3.

Here we assume that the number of mopeds produced = x

and the number of bicycles produced = y

So, the profit equation should be

Profit, P = $134 × x + $20 × y

That provides P = 134x + 20y

4) From Desmos, we used the following constraints;

x + y ≤ 160

10 ≤ x ≤ 60

0 ≤ y ≤ 120 we have;

So, The vertices of the constraint should be

J. (10, 150), I(40, 120), G(60, 100), C(60, 0) and E(10, 0)

5) The maximum no of mopeds should be 60

And, the no of bicycles that need to be produced is

= 160 - 60

= 100 bicycles

That provides  60 mopeds 100 bicycles.

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