2 x 11 multiplies to 22 .
9 + 8 adds to 17 .
(1.4113) and (15.5887) do both jobs.
Answer:
Anything above 15 miles, so x>15
Step-by-step explanation:
When you solve this by using an equation, you would get 77=65+0.80m (0.80 m because slope is 0.80 and m is mileage you multiply by). You would subtract 64 from both sides, so 12=0.80m and then you divide 0.80 by 12 to get m by itself to get 15. 15 miles is when both companies cost the same, so anything above 15.
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
The answer is C i’m pretty sure because you can divide y by 15 and get x
