# How many of each model should you order to minimize the costs? Draw a graph to illustrate your answer. Rest of info written below...You are the manager of a small appliance store. You need to order...

How many of each model should you order to minimize the costs? Draw a graph to illustrate your answer. Rest of info written below...

You are the manager of a small appliance store. You need to order two types of microwave ovens. The deluxe model costs you \$500 and the basic model costs you \$300. The profit for the deluxe model is \$200 while the profit for the basic model is \$50. You would like to sell a minimum of 100 microwaves and make a total profit of at least \$8000.

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

The question makes no practical sense. If the only limitation faced by the manager is the number of ovens sold. it would be best for the manager to buy and sell 100 deluxe model making a total profit of \$20,00 @ \$200 per oven. There is no point in restricting the profit to \$8,000.

However, as given in the question we will assume that the manager has mad a profit of \$8,000 on sale of total of 100 ovens, and we have to find out the number of ovens of each type.

Given:

Profit on deluxe model = \$200 per oven

Profit on basic model = \$50 per oven

Total number of ovens soldĀ  = 100

Let us say the number of deluxe model sold is 'x',

Then:

Number of basic model sold = 100 - x

Then total profit on both the models sold is given by the formula:

x*200 + (100 - x)*50 = 200x + 5000 - 50x = 150x + 5000

But we know total profit is 8000, therefore:

150x + 5000 = 8000

rearranging the terms in the above equation we get:

150x = 8000 - 5000 = 3000

therefor:

x = 3000/150 = 20

Therefore:

Number of deluxe models sold = 20, and

Number of deluxe models sold = 100 - 20 = 80

neela | High School Teacher | (Level 3) Valedictorian

Posted on

The price of the deluxe model = \$500. The profit is \$200 per model.The price of the basic model = \$300 and the profit per model = \$50.

If x and y are the number of models soled, the the constraints are x+y >=100..............(1)

And the profit 200x+50y > \$8000 or 4x+y>160........(2)

From (1) and (2), we get: 4x+100-x >160 or 3x >160-100 =60 Or x>20 and y > 80 , which makes the two condtions satisfy.

To minimise the cost we should minimise, 500x+300y, which is possible only when you keep the higher cost item at minimum and the lower cost to the maximum. This makes a sense of not sell higher cost Delux oven. So x= 0 and y >100 and y >8000/50 =160 .

So the solution for the consrtraints,

x+y>100 and the profit,200x+50y >=8000 and 500x+300y is minumum possible is x=0, y = 160.