Answer to Question #270827 in Operations Research for Rondell Bisnath

Question

A Manufacturer produces two products, the Klunk and the Klick. Klunk has a contribution

to profit of $3, and the Klick $4 per unit. The manufacturer wishes to establish the weekly

production plan that maximizes profit. Production of these products is limited to machine,

labor and material constraints. Each Klunk requires four hours machining, four hours labor

and one kilogram of material, where as each Klick requires two hours machining, six hours

labor and one kilogram of material. Machining and labor has a maximum of one hundred

and one hundred and eighty hours available, and total material available is forty kilograms.

Because of a trade agreement, sales of Klunk are limited to a weekly maximum of twenty

units and to honor an agreement with an old established customer at least ten units of Klick

must be sold each week.

i. Determine graphically using linear programming a suitable production mix of Klunk

and Klick. [12]

ii. What will be the company’s maximum profit?

Answer 1

Let $x=$ the number of units of Klunk to be produced, $y =$the number of units of Klick to be produced.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c} & Machine & Labour & Material \\ & hours & hours & (kg) \\ \hline Klunk & 4x & 4x & x \\ Klick & 2y & 6y & y \\ \hdashline Total & & & \\ available & 100 & 180 & 40 \\ per\ week & & & \\ \end{array}$$

Maximize $z=3x+4y$

Subject to the constraints

$$4x+2y\leq 100$$ $$4x+6y\leq180$$ $$x+y\leq40$$ $$0\leq x\leq20$$ $$y\geq 10$$

i. The solution set of this system is the shaded region in the diagram

$$4x+2y= 100=>y=-2x+50$$ $$x=20, y=10$$

Point $C(20, 10)$

$$4x+6y= 180=>y=-\dfrac{2}{3}x+30$$ $$x=0, y=30$$

Point $A(0, 30)$

$$x=0, y=10$$

Point $D(0, 10)$

$$-2x+50=-\dfrac{2}{3}x+30$$ $$\dfrac{4}{3}x=20$$ $$x=15, y=20$$

Point $B(15, 20)$

ii.

$$\def\arraystretch{1.5} \begin{array}{c:c} Vertices & z=3x+4y \\ \hline A(0,30) & z=3(0)+4(30)=120 \\ \hdashline B(15,20) & z=3(15)+4(20)=125 \\ \hdashline C(20,10) & z=3(20)+4(10)=100 \\ \hdashline D(0,10) & z=3(0)+4(10)=40 \\ \hdashline \end{array}$$

Number of units of Kluck is 15 and number of units of Klick is 20.

Maximum profit is $z(x=15, y=20)=\$125.$