deluxe coffee is to be mixed with regular coffee to make at least 48 pounds of a blended coffee. the mixture mustContain at least 14 pounds of the deluxe coffee. The luxe coffee cost six dollars per pound of regular coffee three dollars per pound how many pounds of each kind of coffee should be used to minimize cost
cost = 6 d + 3 r
d + r = 48
but d = 14 for minimum cost (makes problem trivial)
r = 48 - 14 = 24
To solve this problem, we can use a two-variable linear programming approach. Let's assume:
x = pounds of deluxe coffee
y = pounds of regular coffee
Objective function:
We want to minimize the cost, which can be calculated using the following equation:
Cost = 6x + 3y
Constraints:
1. The total weight of the blended coffee should be at least 48 pounds:
x + y ≥ 48
2. The mixture must contain at least 14 pounds of deluxe coffee:
x ≥ 14
Now we can solve this problem using a graphical method or substitution method:
Graphical Method:
1. Plot the two equations x+y=48 and x=14 on a graph.
2. Identify the feasible region where both constraints are satisfied.
3. Calculate the objective function at the corner points of the feasible region.
4. The corner point with the minimum cost will give us the optimal solution.
Substitution Method:
1. Solve one equation for one variable.
- From the constraint x+y=48, we get y=48-x.
- Substitute this value in the objective function.
2. Minimize the cost by taking the derivative of the objective function with respect to x, set it equal to zero, and solve for x.
3. Calculate the corresponding value of y using y = 48 - x.
Let's use the substitution method:
Substitute y = 48 - x into the objective function:
Cost = 6x + 3(48 - x) = 6x + 144 - 3x = 3x + 144
To minimize the cost, we take the derivative and set it equal to zero:
d(Cost)/dx = 3 = 0
Solving for x:
3x + 144 = 0
3x = -144
x = -48
Since x cannot be negative, there is no valid solution for this problem.
Hence, there is no feasible solution to minimize the cost while meeting the constraints.
To solve this problem, we need to use a technique called linear programming. This technique allows us to determine the optimal solution by minimizing or maximizing an objective function, subject to certain constraints.
Let's define our variables:
Let x be the number of pounds of deluxe coffee used.
Let y be the number of pounds of regular coffee used.
Now, we can set up the objective function and constraints:
Objective Function: Cost = 6x + 3y (We want to minimize the cost.)
Constraints:
1. x + y ≥ 48 (We want at least 48 pounds of blended coffee.)
2. x ≥ 14 (We want at least 14 pounds of deluxe coffee.)
Now, let's graph these constraints on a coordinate plane to visualize the feasible region:
To graph the first constraint x + y ≥ 48, we can rewrite it as y ≥ 48 - x. This gives us the following line:
y = 48 - x
To graph the second constraint x ≥ 14, we draw a vertical line at x = 14.
Now, we shade the region that satisfies both constraints. This shaded region represents all the feasible solutions:
---------|---------|---------|---------|---------
14 20 30 40 48
From the graph, we can see that the feasible region lies to the right and above the line y = 48 - x, and to the right of x = 14.
To find the optimal solution, we need to evaluate the cost function at the corner points of the feasible region.
Let's calculate the cost for each corner point:
1. (14, 34): Cost = 6(14) + 3(34) = 84 + 102 = 186
2. (14, 48): Cost = 6(14) + 3(48) = 84 + 144 = 228
3. (34, 14): Cost = 6(34) + 3(14) = 204 + 42 = 246
From the calculations, we can see that the minimum cost is 186, which occurs at the point (14, 34).
Therefore, to minimize the cost, we should use 14 pounds of deluxe coffee and 34 pounds of regular coffee.