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KiRa [710]
3 years ago
6

Carlos is looking to open an account at a bank, but wants to avoid as many fees as possible. Which would not help minimize fees

paid to a bank?
A. checking with the bank to know what fees it has so he can avoid them

B. only using his bank's ATMs to avoid ATM fees

C. avoiding the use of a check register

D. comparing various banks' fees and services
Business
1 answer:
Nadusha1986 [10]3 years ago
4 0

Answer:

C. avoiding the use of a check register

Explanation:

In order to avoid the extra fee , Carlos can follow the steps as -

1. Consult with the bank to get to know the fees , so as to avoid it .

2. one of the good method , to avoid fees , is to only use ATM of the bank to withdraw cash , so as to avoid the extra charges that are deducted by using other bank's ATM .

3. By comparing the services and fees of all the bank , Carlos , can save extra bucks .

hence , the option , which would not help to reduce the fees is ( C ) .

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Answer:

a. Min Z = 2x₁ + 2.50x₂ + 3x₃

Subject to constraints :

3x₁ + 2x₂ + 4x₃ ≤ 128           .......(1)

3x₁ + x₂ + 3x₃ ≤ 160             ........(2)

x₁ + 0x₂ + 2x₃ ≤ 32              .........(3)

6x₁ + 8x₂ + 4x₃ ≤ 256          .........(4)

3x₁ + 2x₂ + 4x₃ ≥ 64            ..........(5)

3x₁ + x₂ + 3x₃ ≥ 80              ..........(6)

x₁ + 0x₂ + 2x₃ ≥ 16              ...........(7)

6x₁ + 8x₂ + 4x₃ ≥ 128         ............(8)

x₁ ≤ 15                                 ...........(9)

x₂≤ 15                                  ...........(10)

x₃ ≤ 15                                 ...........(11)

x₁ , x₂, x₃ ≥ 0

b. x₁ = 15 ,  x₂ = 9.5 , x₃ = 8.5

c. The optimal solution is Z = 79.25

Explanation:

Given - The table is as follows :

  • Nutrient           Feed 1                    Feed 2                      Feed 3
  •    A                      3                              2                                4
  •    B                      3                               1                                 3
  •    C                      1                                0                                2
  •    D                      6                               8                                4

The minimum requirement per cow each month is 4 pounds of nutrient A, 5 pounds of nutrient B, 1 pound of nutrient C, and 8 pounds of nutrient D. However, cows should not be fed more than twice the minimum requirement for any nutrient each month. Additionally, the ranch can only obtain 1,500 pounds of each type of feed each month. Because there are usually 100 cows at the beef-up ranch at any given time, this means that no more than 15 pounds of each type of feed can be used per cow each month.

To find - a. Formulate a linear programming problem to determine how  

                  much of each type of feed a cow should be fed each month.

              b. Create a spreadsheet model for this problem, and solve it using

                   Solver.

              c. What is the optimal solution?

Proof -

a.

Let feed 1 per cow per month = x₁

     feed 2 per cow per month = x₂

     feed 3 per cow per month = x₃

Now,

As given, The cost per pound of feeds 1,2, and 3 are $2.00, $2.50, and $3.00, respectively.

So, we have to minimize the cost , Z = 2x₁ + 2.50x₂ + 3x₃

Subject to constraints :

3x₁ + 2x₂ + 4x₃ ≤ 4(32)

3x₁ + x₂ + 3x₃ ≤ 5(32)

x₁ + 0x₂ + 2x₃ ≤ 1(32)

6x₁ + 8x₂ + 4x₃ ≤ 8(32)

∴ we get

Min Z = 2x₁ + 2.50x₂ + 3x₃

Subject to constraints :

3x₁ + 2x₂ + 4x₃ ≤ 128           .......(1)

3x₁ + x₂ + 3x₃ ≤ 160             ........(2)

x₁ + 0x₂ + 2x₃ ≤ 32              .........(3)

6x₁ + 8x₂ + 4x₃ ≤ 256          .........(4)

Now, as given

However, cows should not be fed more than twice the minimum requirement for any nutrient each month.

∴ we have

3x₁ + 2x₂ + 4x₃ ≥ \frac{128}{2}

3x₁ + x₂ + 3x₃ ≥ \frac{160}{2}

x₁ + 0x₂ + 2x₃ ≥ \frac{32}{2}

6x₁ + 8x₂ + 4x₃ ≥ \frac{256}{2}

and also

No more than 15 pounds of each type of feed can be used per cow each month.

⇒x₁ , x₂, x₃ ≤ 15

So,

The LPP model becomes

Min Z = 2x₁ + 2.50x₂ + 3x₃

Subject to constraints :

3x₁ + 2x₂ + 4x₃ ≤ 128           .......(1)

3x₁ + x₂ + 3x₃ ≤ 160             ........(2)

x₁ + 0x₂ + 2x₃ ≤ 32              .........(3)

6x₁ + 8x₂ + 4x₃ ≤ 256          .........(4)

3x₁ + 2x₂ + 4x₃ ≥ 64            ..........(5)

3x₁ + x₂ + 3x₃ ≥ 80              ..........(6)

x₁ + 0x₂ + 2x₃ ≥ 16              ...........(7)

6x₁ + 8x₂ + 4x₃ ≥ 128         ............(8)

x₁ ≤ 15                                 ...........(9)

x₂≤ 15                                  ...........(10)

x₃ ≤ 15                                 ...........(11)

x₁ , x₂, x₃ ≥ 0

b.)

We use simplex method calculator  to solve this LPP Problem

we get

x₁ = 15 ,  x₂ = 9.5 , x₃ = 8.5

c.)

The optimal solution is Z = 79.25

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