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.
To find the least number of each type of freezer the manager should order to minimize costs, we can set up a system of equations based on the given information.
Let's say the manager orders x chest freezers and y upright freezers.
The cost of x chest freezers is $250x.
The cost of y upright freezers is $400y.
The profit from x chest freezers is $40x.
The profit from y upright freezers is $60y.
According to the given information, the manager expects to sell at least 100 freezers, so we have the equation:
x + y ≥ 100
The total profit must be at least $4800, so we have the equation:
40x + 60y ≥ 4800
Now, to minimize costs, we need to minimize the total cost, which is the sum of the cost of chest freezers and upright freezers. We can represent this as the objective function:
Cost = 250x + 400y
To solve this problem, we can use linear programming techniques. One approach is the graphical method, where we graph the feasible region and find the minimum value of the objective function within that region.
First, let's graph the two inequality constraints:
x + y ≥ 100 (1)
40x + 60y ≥ 4800 (2)
To graph these equations, we can convert them into slope-intercept form by solving for y:
(1) y ≥ -x + 100
(2) y ≥ -2/3x + 80
Next, we plot the feasible region on a graph, which is the region that satisfies both inequality constraints:
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| (2)
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(0) (100)
Next, we need to find the corner points of the feasible region by solving the equations that define the boundaries of the region.
The corner points of the feasible region are found at the intersections of the two inequality lines, which are (0,100), (60,40), and (100,0).
Now, evaluate the objective function at these corner points:
For (0,100):
Cost = 250(0) + 400(100) = 40000
For (60,40):
Cost = 250(60) + 400(40) = 34000
For (100,0):
Cost = 250(100) + 400(0) = 25000
From the three calculations, we see that the minimum cost occurs at (100,0), where the cost is $25,000. Therefore, the manager should order 100 chest freezers and 0 upright freezers to minimize costs while meeting the profit requirements.
Let's assume the manager orders "x" chest freezers and "y" upright freezers.
The cost of one chest freezer is $250, so the cost of "x" chest freezers is 250x.
The cost of one upright freezer is $400, so the cost of "y" upright freezers is 400y.
The profit from one chest freezer is $40, so the profit from "x" chest freezers is 40x.
The profit from one upright freezer is $60, so the profit from "y" upright freezers is 60y.
The manager expects to sell at least 100 freezers, so we have the equation: x + y ≥ 100.
And the total profit must be at least $4800, so we have the equation: 40x + 60y ≥ 4800.
To minimize costs, we minimize the function: C = 250x + 400y.
Now we can solve this problem using linear programming.
First, we graph the feasible region defined by the two inequalities:
x + y ≥ 100
40x + 60y ≥ 4800
Graphing these equations, we get:
Please wait a moment while I generate the graph for you.
Let C = # of chest freezers and U = # of upright.
40C + 60U ≥ 4800
C + U ≥ 100
Therefore, C ≥ 100 - U.
Substitute 100 - U for C in the first equation to solve for U. Put that value in the second equation to find C. Check by putting both values into the first equation.
I hope this helps.