Answer:
z (min) = 360 x₁ = x₃ = 0 x₂ = 3
Step-by-step explanation:
Protein Carbohydrates Iron calories
Food 1 (x₁) 10 1 4 80
Food 2 (x₂) 15 2 8 120
Food 3 (x₃) 20 1 11 100
Requirements 40 6 12
From the table we get
Objective Function z :
z = 80*x₁ + 120*x₂ + 100*x₃ to minimize
Subjet to:
Constraint 1. at least 40 U of protein
10*x₁ + 15*x₂ + 20*x₃ ≥ 40
Constraint 2. at least 6 U of carbohydrates
1*x₁ + 2*x₂ + 1*x₃ ≥ 6
Constraint 3. at least 12 U of Iron
4*x₁ + 8*x₂ + 11*x₃ ≥ 12
General constraints:
x₁ ≥ 0 x₂ ≥ 0 x₃ ≥ 0 all integers
With the help of an on-line solver after 6 iterations the optimal solution is:
z (min) = 360 x₁ = x₃ = 0 x₂ = 3
Hello :
<span>−4≤16−4x
</span><span>−4 -16 ≤−4x
-20 </span>≤−4x ( muliply by (-1)
4x <span>≥ 20
x</span><span>≥20/4
</span>x≥ 5
Answer:
4!
Step-by-step explanation:
Mode is the number that is repeated the most. In this set of data, the number 4 is represented 6 times which is far more than the other numbers here. Take care!
Answer:
The distance Plane A traveled in 1 h is greater than the distance Plane B traveled in 1 h.
Step-by-step explanation:
The equation of the distance traveled by Plane A is
The plane B traveled 540 miles in 2 hours.
So in 1 hour, plane B traveled 540/2 = 270 miles:
How does the distance Plane A traveled in 1 hour compare to the distance Plane B traveled in 1 hour?
Plane A:
d(1) = 290*1 = 290
Plane A traveled 290 miles in 1 hour.
Plane B travaled 270 miles in 1 hour.
So the correct answer is:
The distance Plane A traveled in 1 h is greater than the distance Plane B traveled in 1 h.
