Answer:
Yes.
Implication : Manipulate demand and choices
Explanation:
<em>Marketing</em> involves communicating the product to the customers at the right price, to the right people and delivering to the right place.
If one of the 4Ps is marketed well for one product customers will have greater attention of that products against another, thus changing the way we think.
Answer:
It is more profitable to continue processing the units.
Explanation:
Giving the following information:
Product A:
Units= 23,000
Selling price= $420,000
Continue processing:
Product B= 6,000 units sold for $106 each
Product C= 11,900 units sold for $52 each
Total cost= $280,000
We need to calculate the effect on the income of both options and choose the most profitable on<u>e. We will not take into account the first costs of Product A because they are irrelevant.</u>
Option 1:
Effect on income= $420,000
Option 2:
Effect on income= (6,000*106) + (11,900*52) - 280,000
Effect on income= $974,800
It is more profitable to continue processing the units.
I believe the correct answer from the choices listed above is the second option. The two <span>participating countries were benefited by global trade in terms of </span><span>economic growth in both the countries. Hope this answers the question. Have a nice day.</span>
Answer:
They should operate Mine 1 for 1 hour and Mine 2 for 3 hours to meet the contractual obligations and minimize cost.
Explanation:
The formulation of the linear programming is:
Objective function:
Restrictions:
- High-grade ore: 
- Medium-grade ore: 
- Low-grade ore: 
- No negative hours: 
We start graphing the restrictions in a M1-M2 plane.
In the figure attached, we have the feasible region, where all the restrictions are validated, and the four points of intersection of 2 restrictions.
In one of this four points lies the minimum cost.
Graphically, we can graph the cost function over this feasible region, with different cost levels. When the line cost intersects one of the four points with the lowest level of cost, this is the optimum combination.
(NOTE: it is best to start with a low guessing of the cost and going up until it reaches one point in the feasible region).
The solution is for the point (M1=1, M2=3), with a cost of C=$680.
The cost function graph is attached.
