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BlackZzzverrR [31]
3 years ago
12

Heidi wants to make some biscuits using this recipe.

Mathematics
1 answer:
Andre45 [30]3 years ago
7 0

The most cookies Heidy can make is 36 cookies.

<h3>How to calculate how many cookies can Heidy make?</h3>

To know how many cookies Heidy can make, you have to take into account the following information:

12 cookies need the following ingredients:

  • 125g butter
  • 200g flour
  • 50g sugar

In the case in which Heidy has more ingredients, we must carry out the following operations:

Divide the quantities, in the reference quantity we have:

  • 500g of butter ÷ 125g of butter = 4
  • 700g flour ÷ 200g flour = 3.5
  • 250g of sugar ÷ 50g of sugar = 5

According to the above, we must take into account the lowest value of all because if that ingredient is enough, we can infer that the rest of the ingredients also.

So the number of cookies Heidy can make are:

12 × 3.5 = 42

Learn more about ingredients in: brainly.com/question/26532763

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3 0
3 years ago
A product can be made in sizes huge, average and tiny which yield a net unit profit of $14, $10, and$5, respectively. Three cent
navik [9.2K]

Answer:

The model is:

z = 14* X₁₁ + 14*X₁₂ + 5*X₁₃ + 10*X₂₁ + 10*X₂₂ + 10*X₂₃ + 5*X₃₁ + 5*X₃₂ + 5*X₃₃    to maximize

Subject to:

First center               X₁₁  +  X₂₁  + X₃₁  ≤  550

Second center         X₁₂  +  X₂₂  + X₃₂  ≤ 750

Third center               X₁₃  + X₂₃  + X₃₃  ≤ 275                  

22* X₁₁  + 16* X₂₁  + 9*X₃₁     ≤   11000

22* X₁₂  + 16* X₂₂  + 9*X₃₂   ≤   2700

22*X₁₃  + 16* X₂₃  +  9*X₃₃  ≤  3400

X₁₁  +  X₁₂  + X₁₃  ≤  710

X₂₁  + X₂₂ + X₂₃  ≤  900

X₃₁ + X₃₂ + X₃₃  ≤  350

2700*(X₁₁  +  X ₂₁  + X₃₁)  -  11000*(X₁₂ + X₂₂ + X₃₂) = 0

3400*(X₁₁  +  X ₂₁  + X₃₁) - 11000*( ( X₁₃ + X₂₃ + X₃₃) = 0

Xij >= 0

Step-by-step explanation:

Let´s call Xij   product size i produced in center j

According to this, we get the following set of variable

X₁₁    product size huge produced in center 1

X₁₂    product size huge produced in center 2

X₁₃   product size huge produced in center 3

X₂₁   product size average produced in center 1

X₂₂   product size average produced in center 2

X₂₃   product size average produced in center 3

X₃₁  product size-tiny produced in center 1

X₃₂ product size-tiny produced in center 2

X₃₃ product size-tiny produced in center 3

Then Objective function is

z = 14* X₁₁ + 14*X₁₂ + 5*X₁₃ + 10*X₂₁ + 10*X₂₂ + 10*X₂₃ + 5*X₃₁ + 5*X₃₂ + 5*X₃₃

Constrains

Center capacity

1.-   First center               X₁₁  +  X₂₁  + X₃₁  ≤  550

2.-   Second center         X₁₂  +  X₂₂  + X₃₂  ≤ 750

3.- Third center               X₁₃  + X₂₃  + X₃₃  ≤ 275

Water available

1.-  22* X₁₁  + 16* X₂₁  + 9*X₃₁     ≤   11000

2.-  22* X₁₂  + 16* X₂₂  + 9*X₃₂   ≤   2700

3.-   22*X₁₃  + 16* X₂₃  +  9*X₃₃  ≤  3400

Demand constrain

Product huge

X₁₁  +  X₁₂  + X₁₃  ≤  710

Product average

X₂₁  + X₂₂ + X₂₃  ≤  900

Product tiny

X₃₁ + X₃₂ + X₃₃  ≤  350

Fraction SP/CC must be the same

First and second centers  fraction SP/CC  

(X₁₁  +  X ₂₁  + X₃₁)/ 11000   =  (X₁₂ + X₂₂ + X₃₂)/ 2700

2700*(X₁₁  +  X ₂₁  + X₃₁)  -  11000*(X₁₂ + X₂₂ + X₃₂) = 0

First and third centers  fraction SP/CC  

(X₁₁  +  X ₂₁  + X₃₁)/ 11000   = ( X₁₃ + X₂₃ + X₃₃)/ 3400

3400*(X₁₁  +  X ₂₁  + X₃₁) - 11000*( ( X₁₃ + X₂₃ + X₃₃) = 0

The model is:

z = 14* X₁₁ + 14*X₁₂ + 5*X₁₃ + 10*X₂₁ + 10*X₂₂ + 10*X₂₃ + 5*X₃₁ + 5*X₃₂ + 5*X₃₃

Subject to:

First center               X₁₁  +  X₂₁  + X₃₁  ≤  550

Second center         X₁₂  +  X₂₂  + X₃₂  ≤ 750

Third center               X₁₃  + X₂₃  + X₃₃  ≤ 275                  

22* X₁₁  + 16* X₂₁  + 9*X₃₁     ≤   11000

22* X₁₂  + 16* X₂₂  + 9*X₃₂   ≤   2700

22*X₁₃  + 16* X₂₃  +  9*X₃₃  ≤  3400

X₁₁  +  X₁₂  + X₁₃  ≤  710

X₂₁  + X₂₂ + X₂₃  ≤  900

X₃₁ + X₃₂ + X₃₃  ≤  350

2700*(X₁₁  +  X ₂₁  + X₃₁)  -  11000*(X₁₂ + X₂₂ + X₃₂) = 0

3400*(X₁₁  +  X ₂₁  + X₃₁) - 11000*( ( X₁₃ + X₂₃ + X₃₃) = 0

Xij >= 0

6 0
3 years ago
2/3n + 5 greater than 12
Tanzania [10]
2/3n + 5 > 12
2/3n > 12 - 5
2/3n > 7
n > 7 * 3/2
n > 21/2 or 10 1/2
6 0
3 years ago
Read 2 more answers
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