Answer:
z(max) = 256000 Php
x₁ = 10
x₂ = 110
Step-by-step explanation:
Jogging pants design Selling Price Cost
weekly production
Design A x₁ 2500 1750
Design B x₂ 2100 1200
1. z ( function is : )
z = 2500*x₁ + 2100*x₂ to maximize
First constraint weekly production
x₁ + x₂ ≤ 120
Second constraint Budget
1750*x₁ + 1200*x₂ ≤ 150000
Then the model is
z = 2500*x₁ + 2100*x₂ to maximize
Subject to
x₁ + x₂ ≤ 120
1750*x₁ + 1200*x₂ ≤ 150000
General constraints x₁ ≥ 0 x₂ ≥ 0 both integers
First table
z x₁ x₂ s₁ s₂ cte
1 -2500 -2100 0 0 0
0 1 1 1 0 = 120
0 1750 1200 0 1 = 150000
Using AtoZmath online solver and after 6 iterations the solution is:
z(max) = 256000 Php
x₁ = 10
x₂ = 110
Answer:
I believe it would be C.
Step-by-step explanation:
Increase $110,000 by 20% 3 times.
110,000 x 0.2 = 22000
(0.2 is 20% as a decimal, we needed to convert it to multiply it)
So we must add 22000 to 110,000, then take 20% of the new cost, and repeat.
Add them
110,000 + 22000 = 132000
Take 20% of new value
132000 x 0.2 = 26400
Add that
132000 + 26400 = 158400
Take another 20% of that
31680
Add them
158400 + 31680 = 190080
So the value is now $190,080
A much more efficient way to do this would be multiplying 1.2 instead of 0.2, and skipping the adding part, as you already took 100% of it and are adding 20% more.
Hope this helps!
Answer:
2nd is the correct answer for your question
Answer:
B. (12-6) n - 50
Step-by-step explanation:
Given that:
Cost of making hat = $6 per hat
Selling price = $12 per hat
Advert cost = $50
Number of hats made = n
The profit or money raised from the sales of hat:
Total Revenue from sales - total cost of production - advert cost
Total Revenue from sales = ($12 * n) = $12n
Total cost of production = ($6 * n) = $6n
Advert cost = $50
Hence,
12n - (6n + 50)
12n - 6n - 50
(12 - 6)n - 50
