Polygon of constraints L.O. graph the polygon of constraints   This is the THIRD part of the optimization problem. When you graph the inequalities the part you shade will form a POLYGON (shape). The goal in this part is to find the VERTICES (corners) of the shape. Some vertices you will be able to read off the graph but others you will have to solve ALGEBRAICALLY . We solve for the vertices algebraically by solving the SYSTEM OF EQUATIONS of the two lines.

Polygon of constraints L.O. graph the polygon of constraints Example: Find the vertices of the following polygon of constraints   𝑥≥0 𝑦≥0 20𝑥+𝑦≤500 𝑦<300 Step 1: Graph the system of inequalities L1: 20𝑥+𝑦≤500 𝑥=0 𝑦=500 𝑦=0 20𝑥=500 𝑥= 500 20 𝑥=25 𝑇𝑒𝑠𝑡 𝑝𝑜𝑖𝑛𝑡 0,0 0+0≤500 0≤500 True x y 500 25

Polygon of constraints L.O. graph the polygon of constraints Example: Find the vertices of the following polygon of constraints   𝑥≥0 𝑦≥0 20𝑥+𝑦≤500 𝑦<300 Step 1: Graph the system of inequalities x y 300 20 L2: 𝑦<300 𝑦=300 Horizontal line 𝑇𝑒𝑠𝑡 𝑝𝑜𝑖𝑛𝑡 0,0 0<300 𝑇𝑟𝑢𝑒

Polygon of constraints L.O. graph the polygon of constraints

Polygon of constraints L.O. graph the polygon of constraints Step 2: Identify the vertices and write the coordinates by either reading it off the graph or solve algebraically 𝐴 0,0 𝐵 0,300 𝐷 25,0 Vertex C looks like it is 10,300 . We will solve it algebraically to make sure we are correct. 𝐿1:20𝑥+𝑦=500 𝐿2:𝑦=300 Use Substitution Method: 20𝑥+300=500 20𝑥=500−300 20𝑥=200 𝑥= 200 20 𝑥=10 →𝐶 10,300

Polygon of constraints L.O. graph the polygon of constraints Step 3: Answer 𝐴 0,0 𝐵 0,300 𝐷 25,0 𝐶 10,300