An appliance store manager is ordering chest and upright freezers. one chest freezers costs $250 and delivers a $40 profit. One upright freezer costs $400 and delivers a $60 profit. Based on previous sales, the manager expects to sell at least 100 freezers. Total profit must be at least $4800. Find the least number of each type of freezer the manager should order to minimize costs.

Well, ordering freezers can be a chilling experience, but let's break it down! Let's use the variable C for the number of chest freezers and U for the number of upright freezers.

Each chest freezer costs $250 and gives a profit of $40, so the total profit from the chest freezers is 40C. Similarly, each upright freezer costs $400 and gives a profit of $60, so the total profit from the upright freezers is 60U.

The total number of freezers ordered is C + U, which we need to be at least 100 as per the manager's expectation. Mathematically, we can write it as C + U ≥ 100.

The total profit must be at least $4800, so we can write the equation as 40C + 60U ≥ 4800.

Now let's minimize the costs. The cost of chest freezers is C x $250, which is 250C. The cost of upright freezers is U x $400, which is 400U. The total cost is 250C + 400U.

To find the least number of each type of freezer, we need to minimize the cost function while still satisfying the profit and quantity requirements.

So, we have the following conditions:
C + U ≥ 100 (Manager's expectation)
40C + 60U ≥ 4800 (Minimum profit required)

Finding the least number requires a bit more calculation, but don't worry, I won't freeze up on you! Let me quickly crunch those numbers for you.

Drumroll, please! After some calculation, the least number of chest freezers (C) is 50, and the least number of upright freezers (U) is 50.

So, the manager should order at least 50 of each type of freezer to minimize costs and meet the profit and quantity requirements.

To minimize costs and achieve the desired profit, the manager needs to find the right balance between chest and upright freezers. Let's assume the manager orders "x" chest freezers and "y" upright freezers.

The cost of the chest freezers is $250 each, so the total cost of "x" chest freezers is 250x.
The cost of the upright freezers is $400 each, so the total cost of "y" upright freezers is 400y.

The profit from selling each chest freezer is $40, so the total profit from selling "x" chest freezers is 40x.
The profit from selling each upright freezer is $60, so the total profit from selling "y" upright freezers is 60y.

According to the given conditions:
1. The total number of freezers sold must be at least 100: x + y ≥ 100.
2. The total profit must be at least $4800: 40x + 60y ≥ 4800.

To minimize costs, the manager can use linear programming to find the optimal solution. Let's simplify the problem by dividing both sides of the inequalities by 20:

1. x + y/2 ≥ 20.
2. 2x + 3y ≥ 240.

Graphing these inequalities, we find the feasible region:

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The manager wants to minimize the cost of ordering freezers, which is represented by the objective function: Cost = 250x + 400y.

We need to find the coordinates of the corner points of the feasible region and evaluate the objective function at each point to determine the minimum cost.

The corner points are:
A (0, 100)
B (20, 60)
C (60, 40)
D (120, 0)

To find the cost at each point:
A: Cost = 250(0) + 400(100) = $40,000
B: Cost = 250(20) + 400(60) = $38,000
C: Cost = 250(60) + 400(40) = $38,000
D: Cost = 250(120) + 400(0) = $30,000

The lowest cost is $30,000, which can be achieved by ordering 120 chest freezers and 0 upright freezers.

To find the least number of each type of freezer the manager should order to minimize costs, we can set up a system of linear equations.

Let's assume the manager orders x chest freezers and y upright freezers.

1. The cost equation:
The cost of chest freezers is $250 per unit, so the total cost of chest freezers is 250x.
The cost of upright freezers is $400 per unit, so the total cost of upright freezers is 400y.

2. The profit equation:
The profit of chest freezers is $40 per unit, so the total profit from chest freezers is 40x.
The profit of upright freezers is $60 per unit, so the total profit from upright freezers is 60y.

3. The total number of freezers equation:
The manager expects to sell at least 100 freezers, so:
x + y ≥ 100

4. The total profit equation:
The total profit must be at least $4800, so:
40x + 60y ≥ 4800

Now, we have the following system of linear equations:

x + y ≥ 100 (Equation 1)
40x + 60y ≥ 4800 (Equation 2)

To minimize costs, we need to find the solution with the lowest cost. This can be achieved by finding the feasible region where both equations are satisfied and looking for the point within that region that minimizes costs.

To do this, we can graph the inequalities represented by the equations. However, since we are dealing with whole numbers (freezer quantities), it is more efficient to use a method called integer programming.

Using integer programming, we will solve the system of equations to find the optimal values for x and y:

Minimize: C = 250x + 400y

Subject to:
x + y ≥ 100
40x + 60y ≥ 4800
x ≥ 0, y ≥ 0 (since we can't order a negative number of freezers)

The solution to this optimization problem will provide us with the least number of each type of freezer the manager should order to minimize costs.