A manufacturer of projection TVs must ship a total of at least 1000 TVs to its two central warehouses, x to the first warehouse and y to the second warehouse. Each warehouse can hold a maximum of 750 TVs. The first warehouse already has 150 TVs on hand, whereas the second has 50 TVs on hand. It costs $6 to ship a TV to the first warehouse, and it costs $17 to ship a TV to the second warehouse. How many TVs should be shipped to each warehouse to minimize cost?
(x, y) =
What is the minimum cost?
follow this procedure ....
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you want to minimize c = 6x+17y
subject to
x+y >= 1000
x <= 750-150
y <= 750-50
now graph the region and evaluate c at the vertices.
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To minimize cost, we need to ship the TVs in a way that minimizes the total amount spent. Let's break down the problem step by step.
We have two warehouses: the first can hold 750 TVs and already has 150, while the second can hold 750 TVs and already has 50.
Let's say we ship x TVs to the first warehouse and y TVs to the second warehouse. We know that x + y must be at least 1000 (the minimum number of TVs required).
x + y ≥ 1000
Since each warehouse has a maximum capacity of 750, we also need to consider those limits:
x ≤ 750 - 150 = 600 (remaining capacity in the first warehouse)
y ≤ 750 - 50 = 700 (remaining capacity in the second warehouse)
Now, we need to minimize the cost. Shipping a TV to the first warehouse costs $6, while shipping it to the second warehouse costs $17.
The total cost (C) can be calculated as:
C = 6x + 17y
To find the minimum cost, we need to find the values of x and y that satisfy the constraints and minimize the cost function.
To optimize this problem, we will use linear programming techniques. By solving the constraints and cost function, we will find the minimum cost and corresponding values of x and y.
Let me do the math for you.
To minimize cost, we need to set up and solve a system of linear inequalities.
Let's define the following variables:
x = the number of TVs shipped to the first warehouse
y = the number of TVs shipped to the second warehouse
We need to determine the values of x and y that satisfy the following conditions:
1. The total number of TVs shipped must be at least 1000:
x + y ≥ 1000
2. The first warehouse can hold a maximum of 750 TVs:
x ≤ 750
3. The second warehouse can hold a maximum of 750 TVs:
y ≤ 750
4. The first warehouse already has 150 TVs on hand:
x + 150 ≤ 750 --> x ≤ 600
5. The second warehouse already has 50 TVs on hand:
y + 50 ≤ 750 --> y ≤ 700
Now, we can solve this system of inequalities to find the optimal solution.
The feasible region is bounded by the lines x + y = 1000, x = 750, y = 750, x = 600, and y = 700. It forms a polygonal region within these constraints.
To minimize the cost, we need to minimize the cost function, which is given by:
Cost = 6x + 17y
We evaluate the cost function at each corner point of the feasible region and determine the minimum value.
Corner points (x, y) to evaluate:
A: (600, 700)
B: (600, 750)
C: (750, 700)
Calculating the cost at each corner point:
Cost(A) = 6(600) + 17(700) = 10,200
Cost(B) = 6(600) + 17(750) = 11,100
Cost(C) = 6(750) + 17(700) = 13,050
The minimum cost is achieved at point A with Cost(A) = $10,200.
Therefore, in order to minimize cost, 600 TVs should be shipped to the first warehouse and 700 TVs should be shipped to the second warehouse. The minimum cost is $10,200.