Jessie’s Jewelry Name: Jessie Rich

Business plan In my jewelry store I have two types of watches that I am selling. One is Calvin Klein and the other is Casio. The Calvin Klein watches are being sold at $350 and the Casio watches are $140.

Variables and constraints Variables: x = Calvin Klein watches y = Casio watches Constraints: There is a budget of $25,000. The Calvin Klein watches cost me $250 from the vendor and the Casios cost me $90. I have to have 10 of both in stock (so, order at least 10 of each), but I can only order up to 50 of the Calvin Klein watches at a time. Lastly, I have to buy at least two of the Casios for every one Calvin Klein that I buy.

Profit Function P = 100x + 50y (I sell Calvin Klein watches at $100 more than what it costs me to buy them from the vendor, and the Casios at $50 more.)

Equations x ≥ 10 y ≥ x + 90y ≤ 25,000 y ≥ 2x x ≤ 50

Graph

Algebra

Possible “best” solutions (10, 20) = 10 Calvin and 20 Casios (50, 100) = 50 Calvin and 100 Casios (10, 250) = 10 Calvin and 250 Casios (50, ) = 50 Calvin and 138 Casios (Can’t have a fraction of a watch)

Best solution- Using P=100x+50y (10,20)- P=100(10)+50(20) = $2,000 (50, 100)- P=100(50)+50(100)= $10,000 (50, 138)- P=100(50)+50(138)= $11,900 (10, 250)- P=100(10)+50(250)= $13,500 Assuming all watches sold, the way to make the most profit would be to buy only 10 Calvin Klein watches and 250 Casio watches.