Question

A company produces two types of tables, $T_1$ and $T_2$. It takes $2$ hours to produce the parts of one unit of $T_1$, $1$ hour to assemble and $2$ hours ot polish. It takes $4$ hours to produce the parts of one unit of $T_2$, $2.5$ hour to assemble and $1.5$ hours of polish. Per month, $7000$ hours are available for producing the parts, $4000$ hours for assembling the parts and $5500$ hours for polishing the tables. The profit per unit of $T_1$ is 90$andperunitof${T}_{2}$is$ $110$. How many of each type of tables should be produced in order to maximize the total monthly profit?

Answer 1

Let $x$ be the number of tables of type $T_1$ and $y$ the number of tables of type $T_2$. Profit $P(x,y)=90x+110y$
$\{\begin{array}{c}x\ge 0\\ x\ge 0\\ 2x+4y\le 7000\\ x+2.5y\le 4000\\ 2x+1,5y\le 5500\end{array}$
The solution set of the system of inequalities above the vertices of the feasible solution set obtained are shown below.
$A$ at $(0,0)$
$B$ at $(0,1600)$
$C$ at $(1500,1000)$
$D$ at $(2300,600)$
$E$ at $(2750,0)$
Evaluate profit $P(x,y)$ at each vertex
$A$ at $(0,0):P(0,0)=0$
$B$ at $\left(0.1600\right):P(0,1600)=90\left(0\right)+110\left(1600\right)=176000$
$C$ at $(1500,1000):P(1500,1000)=90\left(2500\right)+110\left(1000\right)=245000$
$D$ at $(2300,600):P(2300,600)=90\left(2300\right)+110\left(600\right)=273000$
$E$ at $(2750,0):P(2750,0)=90\left(2750\right)+110\left(0\right)=247500$
The maximum profit of 273000$isatvertex$D$.Hencethecompanyneedstoproduct$2300$tablesoftype${T}_{1}$and$600$tablesoftype${T}_{2}$ in order to maximize its profit.