6.1 Method, exercises

Method: linear programming

If we wish to maximise the objective function $f(x,y)$ then:

1. Find the gradient of the level lines of $f(x,y)$ (this is always going to be $-\frac{a}{b}$ as we saw in Equation  [link] )
2. Place your ruler on the $xy$ plane, making a line with gradient $-\frac{a}{b}$ (i.e. $b$ units on the $x$ -axis and $-a$ units on the $y$ -axis)
3. The solution of the linear program is given by appropriately moving the ruler. Firstly we need to check whether $b$ is negative, positive or zero.
   1. If $b>0$ , move the ruler up the page, keeping the ruler parallel to the level lines all the time, until it touches the “highest” point in the feasible region. This point is then the solution.
   2. If $b<0$ , move the ruler in the opposite direction to get the solution at the “lowest” point in the feasible region.
   3. If $b=0$ , check the sign of $a$
      1. If $a<0$ move the ruler to the “leftmost” feasible point. This point is then the solution.
      2. If $a>0$ move the ruler to the “rightmost” feasible point. This point is then the solution.

End of chapter exercises

1. Polkadots is a small company that makes two types of cards, type X and type Y. With the available labour and material, the company can make not more than 150 cards of type X and not more than 120 cards of type Y per week. Altogether they cannot make more than 200 cards per week. There is an order for at least 40 type X cards and 10 type Y cards per week.Polkadots makes a profit of R5 for each type X card sold and R10 for each type Y card. Let the number of type X cards be x and the number of type Y cards be y, manufactured per week.
   1. One of the constraint inequalities which represents the restrictions above is $x\le 150$ . Write the other constraint inequalities.
   2. Represent the constraints graphically and shade the feasible region.
   3. Write the equation that represents the profit $P$ (the objective function), in terms of $x$ and $y$ .
   4. On your graph, draw a straight line which will help you to determine how many of each type must be made weekly to produce the maximum $P$
   5. Calculate the maximum weekly profit.
2. A brickworks produces “face bricks" and “clinkers". Both types of bricks are produced and sold in batches of a thousand. Face bricks are sold at R150 per thousand, and clinkers at R100 per thousand, where an income of at least R9,000 per month is required to cover costs. The brickworks is able to produce at most 40,000 face bricks and 90,000 clinkers per month, and has transport facilities to deliver at most 100,000 bricks per month. The number of clinkers produced must be at least the same number of face bricks produced. Let the number of face bricks in thousands be $x$ , and the number of clinkers in thousands be $y$ .
   1. List all the constraints.
   2. Graph the feasible region.
   3. If the sale of face bricks yields a profit of R25 per thousand and clinkers R45 per thousand, use your graph to determine the maximum profit.
   4. If the profit margins on face bricks and clinkers are interchanged, use your graph to determine the maximum profit.
3. A small cell phone company makes two types of cell phones: Easyhear and Longtalk . Production figures are checked weekly. At most, 42 Easyhear and 60 Longtalk phones can be manufactured each week. At least 30 cell phones must be produced each week to cover costs. In order not to flood the market, the number of Easyhear phones cannot be more than twice the number of Longtalk phones. It takes $\frac{2}{3}$ hour to assemble an Easyhear phone and $\frac{1}{2}$ hour to put together a Longtalk phone. The trade unions only allow for a 50-hour week. Let $x$ be the number of Easyhear phones and $y$ be the number of Longtalk phones manufactured each week.
   1. Two of the constraints are:
      $$0\le x\le 42\mathrm{and}0\le y\le 60$$
      Write down the other three constraints.
   2. Draw a graph to represent the feasible region
   3. If the profit on an Easyhear phone is R225 and the profit on a Longtalk is R75, determine the maximum profit per week.
4. Hair for Africa is a firm that specialises in making two kinds of up-market shampoo, Glowhair and Longcurls . They must produce at least two cases of Glowhair and one case of Longcurls per day to stay in the market. Due to a limited supply of chemicals, they cannot produce more than 8 cases of Glowhair and 6 cases of Longcurls per day. It takes half-an-hour to produce one case of Glowhair and one hour to produce a case of Longcurls , and due to restrictions by the unions, the plant may operate for at most 7 hours per day. The workforce at Hair for Africa , which is still in training, can only produce a maximum of 10 cases of shampoo per day. Let $x$ be the number of cases of Glowhair and $y$ the number of cases of Longcurls produced per day.
   1. Write down the inequalities that represent all the constraints.
   2. Sketch the feasible region.
   3. If the profit on a case of Glowhair is R400 and the profit on a case of Longcurls is R300, determine the maximum profit that Hair for Africa can make per day.
5. A transport contracter has 6 5-ton trucks and 8 3-ton trucks. He must deliver at least 120 tons of sand per day to a construction site, but he may not deliver more than 180 tons per day. The 5-ton trucks can each make three trips per day at a cost of R30 per trip, and the 3-ton trucks can each make four trips per day at a cost of R120 per trip. How must the contracter utilise his trucks so that he has minimum expense?