1. Matrices Task

2. The Scenario
A florist has to order roses and carnations for Valentine's Day. The florist needs to decide how many dozens of roses and carnations should be ordered to obtain a maximum profit.
Roses: The florist's cost is $20 per dozen; the profit over cost is $20 per dozen.
Carnations: The florist's cost is $5 per dozen; the profit over cost is $8 per dozen.
The florist can order no more than 60 dozen flowers.
A minimum of 20 dozen carnations must be ordered.
The florist cannot order more than $450 worth of roses and carnations.

3. The Solutions
The equations to represent the scenario would be: X + Y <=60, where Y >= 20 20X + 5 Y <=$450 The work: 1200 – 20y + 5y <= y <= 450 Y <=50, therefore X >=10

4. Graph The X axis represents roses, Y axis is carnations.
The maximum profit is the intersection at point (10, 50)
The florist should buy 10 dozen roses and 50 dozen carnations to maximize profits.

5. The Reflection
I selected option one which was to assist a florist in determining how to maximize their profits for Valentine’s Day. I chose this scenario because I enjoy the look and smell of flowers and was curious how profitable it was to sell roses on Valentine’s Day. Most women prefer roses over carnations, but I know most men don’t want to spend the money it takes to buy roses. I was curious from the florists point of view what they thought was best from a business standpoint. To best display my scenario, I chose to determine the equations for the matrices and the graph out the solution to show the maximum objective of the scenario. The graph is the best picture to show what the possibilities are that solve for both equations. My answer for the florist was to buy 50 dozen carnations and 10 dozen roses. So to conclude from the florist point of view, they would rather sell carnations over roses at a 5 to 1 ratio because they allow for the most profits in this scenario. The total profit would be $600 if the florist bought and sold the 50 dozen carnations and 10 dozen roses.