12/y=4/5 so 4y/5=12 so then 4y=60 and y=15
x/5=y/3 so x/5=15/3 so x/5=5 so x=25
Answer:
D
Step-by-step explanation:
w ≥ 15
Answer:
1/7
Step-by-step explanation:
x/y
Plug in the values.
(2)/(14)
Simplify.
1/7
Answer:
To complete the problem statement it is needed:
1.- the volume and weight capacity of the truck, because these will become the constraints.
2.- In order to formulate the objective function we need to have an expression like this:
" How many of each type of crated cargo should the company shipped to maximize profit".
Solution:
z(max) = 175 $
x = 1
y = 1
Assuming a weight constraint 700 pounds and
volume constraint 150 ft³ we can formulate an integer linear programming problem ( I don´t know if with that constraints such formulation will be feasible, but that is another thing)
Step-by-step explanation:
crated cargo A (x) volume 50 ft³ weigh 200 pounds
crated cargo B (y) volume 10 ft³ weigh 360 pounds
Constraints: Volume 150 ft³
50*x + 10*y ≤ 150
Weight contraint: 700 pounds
200*x + 360*y ≤ 700
general constraints
x ≥ 0 y ≥ 0 both integers
Final formulation:
Objective function:
z = 75*x + 100*y to maximize
Subject to:
50*x + 10*y ≤ 150
200*x + 360*y ≤ 700
x ≥ 0 y ≥ 0 integers
After 4 iterations with the on-line solver the solution
z(max) = 175 $
x = 1
y = 1
Can you please elaborate on your question