let x = the # of cans…

let y = the # of chairs…

Your two inequalities are…

Assembly center limitation: x + 1.5y <= 12 (Equation 1) Finishing center limitation: 1.33x + 1.5y <= 15 (Equation 2) Those two equations are the two equations of the lines that you are going to plot on graph paper. But FIRST you have to change them into the general line equation format of y = mx + b.... So Equation 1... x + 1.5y <= 12 becomes... ** subtract "x" from both sides... and you get... ** 1.5y <= -x + 12 now divide both sides by "1.5" (which is the same as multiplying throug by "2/3".... and you get... y = -(2/3)x + 12(2/3) y = -(2/3)x + 8 (Line 1 in y = mx + b format) **LINE 1** ----------------------- Now let's change Equation 2 to y = mx + b format So Equation 2... 1.33x + 1.5y <= 15 becomes... ** subtract "1.33x" from both sides... and you get... ** 1.5y <= -1.33x + 15 .... -1.33 is the same as 1 1/3 = 4/3 now divide both sides by "1.5" (which is the same as multiplying throug by "2/3".... and you get... y = -(2/3)(1.33)x + 15(2/3) y = -(2/3)(4/3)x + 15(2/3) y = -(8/9)x +10 (Line 2 in y = mx + b format) **LINE 2** ------------------------------------------ RECAP: This is what we found... A) First we found the two inequalities... Assembly center limitation: x + 1.5y <= 12 (Equation 1) Finishing center limitation: 1.33x + 1.5y <= 15 (Equation 2) B) Then we changed them to y = mx + b format y = -(2/3)x + 8 (Line 1 in y = mx + b format) **LINE 1** y = -(8/9)x +10 (Line 2 in y = mx + b format) **LINE 2** -------------------------------------------- Now I assume you know how to plot lines.... So plot LINE 1..... after you plot it... "shade" with / / / / / / hatching the whole area BELOW that line.... Okay.... now plot LINE 2 ..... after you plot the 2nd line... "shade" with \ \ \ \ \ hatching the whole area BELOW the 2nd line.... Okay.... no you should see an area where the / / / / / and the \ \ \ \ intersect forming diamonds... or X X X X X X hatching... Do you see that? It will look like fish netting. THAT area where you see fish netting/diamonds.... is your solution..... those are ALL the possible production levels that can occur that will meet the requirements of BOTH inequalities.... The area where you ONLY have / / / / / hatching.... and the area where you ONLY have \ \ \ \ \ will NOT meet BOTH the limitations of the assembly center and the finishing center. The area where you ONLY have / / / / / hatching is NOT a solution to this problem.... it is only a solution to meeting the requirements of the assembly center. The area where you ONLY have \ \ \ \ \ hatching is NOT a solution to this problem.... it is only a solution to meeting the requirements of the finishing center. Only the area where you see XXXXX XXXXX XXXXX hatching.... that is the solution to this problem.... Hope this helps! It's hard to graph in this forum so I can only describe the steps to you as far as how to set up the problem to find the solution....