Answer:
z - 2*x - 1.5*y = 0 maximize
subject to:
3*x + 5*y ≤ 800
8*x + 3*y ≤ 1200
x, y > 0
Step-by-step explanation:
Formulation:
Kane Manufacturing produce x units of model A (fireplace grates)
and y units of model B
quantity Iron cast lbs labor (min) Profit $
Model A x 3 8 2
Model B y 5 3 1.50
We have 800 lbs of iron cast and 1200 min of labor available
We need to find out how many units x and units y per day to maximiza profit
First constraint Iron cast lbs 800 lbs
3*x + 5*y ≤ 800 3*x + 5*y + s₁ = 800
Second constraint labor 1200 min available
8*x + 3*y ≤ 1200 8*x + 3*y + s₂ = 1200
Objective function
z = 2*x + 1.5*y to maximize z - 2*x - 1.5*y = 0
x > 0 y > 0
The first table is ( to apply simplex method )
z x y s₁ s₂ Cte
1 -2 -1.5 0 0 0
0 3 5 1 0 800
0 8 3 0 1 1200
5x^2 + 20
Option B is the answer
5(x^2+4)
(5 * x^2) + (5 * 4)
5x^2 + 20
Total number of students in the class = 36
Number of students wearing blue shirts = (1/6) * 36
= 36/6
= 6
Number of students wearing white shirts = (2/3) * 36
= 2 * 12
= 24
Then
Number of students not wearing blue or white shirts in the class = 36 - (6 + 24)
= 36 - 30
= 6
So 6 students in the class are wearing a shirt other than a blue or a white shirt. I hope the procedure is clear enough for you to understand.
It might be 7ab
Hope this helps!
