Linear programming graphically

A small firm produces two types of wooden lampstands: rounded and angular. Both types require two hand-crafted processes: cutting and smoothing. Rounded lampstands require 1 hour of cutting and 3 hours of smoothing whereas angular lampstands require 2 hours of cutting but only 1 hour of smoothing. The firm has 400 man-hours of cutting available each week and 300 man-hours of smoothing. The firm calculates that they can make 3 Euro profit on each rounded lampstand and 4 Euro profit on each angular lampstand. The problem is to maximise profit. Formulate this as a linear programming problem, giving the three steps and state any assumptions made. (b) Solve the LP problem in (a) graphically. Hence, state your recommendation for the number of each type of lampstands the firm should produce in order to maximise weekly profit. Give the total profit per week that would be expected given your solution.

How do I draw it graphically and how much profit will they be expected to make in a week? Hope you can help. Thank You

Let $x$ be the number of rounded lampstands and $y$ be the number of angular lampstands. Then we see that our profit is $P(x,y) = 3x+4y$ subject to the following constraints: $$x \geq 0$$ $$y \geq 0$$ $$x+2y \leq 400 \text{ (cutting)}$$ $$3x+y \leq 300 \text{ (smoothing)}$$

Graphing these, we get the following quadrilateral as our feasibility region: We know that our solution must be one of the vertices of the polygon formed by the feasibility region, thus, we test them all: $$P(0,0) = 3(0)+4(0) = \0$$ $$P(0,200) = 3(0)+4(200) = \800$$ $$P(100, 0) = 3(100)+4(0) = \300$$ $$P(40,180) = 3(40)+4(180) = \840$$

Thus, we see it is best to make $40$ rounded lampstands and $180$ angular lampstands in order to maximize profit.

• Thanks. can you show me how you got the points: (0,200), (0,0), (40,180) and (100,0). Thanks – John Aug 13 '16 at 11:46
• If you graph all the inequalities and see where all the shaded regions intersect, you get the quadrilateral formed by those 4 points. The origin is somewhat obvious (gotten from $x>0$ and $y=0$). Notice the top-left point is gotten by seeing where the line $x+2y=400$ hits $y$, i.e., $x=0$ which implies $y=200$. Similarly, we need the line $3x+y=300$ to hit the $x$-axis, giving us $y=0$ and $x=100$. The last point is gotten by seeing where the two lines intersect. If you subject the equation for the 2nd line from 3 times the 1st equation, you get $3(2y) -y = 3(400)-300$, thus, $y= 180$. – benguin Aug 13 '16 at 11:57
• Substitute $y=180$ into one of the equations for the two lines to get $x=40$. – benguin Aug 13 '16 at 11:59