Question

a) A manufacturing company makes two models $A$ and $B$ of a product. Each piece of model $A$ requires $9$ labor hours for fabricating and $1$ labor hour for finishing. Each piece of model $B$ requires $12$ labor hours for finishing and $3$ labor hours for finishing. For fabricating and finishing, the maximum labor hours available are: $180$ and $30$ respectively. The company makes a profit of Rs. $8000$ on each piece of model $A$ and Rs. $12000$ on each piece of model $B$. How many pieces of model $A$ and model $B$ should be manufactured per week to realize a maximum profit? What is the maximum profit per week?
b) Find the value of $K$ so that the function $f\left(x\right)=\begin{array}{c}\{\begin{array}{cc}Kx+1,& \text{if}x\le 5\\ 3x-5,& \text{if}x\le 5\end{array}\end{array}$ at $x=5$ is a continuous function.

Answer 1

$\left(a\right)$

Let $x$ be the number of pieces of model $A$ and $y$ be the number of pieces of model $B$.

Let total profit be $Z$

Then, Total profit $=8000x+12000y$

$Z=8000x+12000y$ .............. $\left(i\right)$

Now we have the following mathematical model for the given problem

Max $Z=8000x+12000y$

Subject to constraints

$9x+12y\le 180$ .......... (Fabricating constraints)

$3x+4y\le 60$ .......... $\left(ii\right)$

$x+3y\le 30$ (Finishing constraints)........ $\left(iii\right)$

$x\ge 0,y\ge 0$ (Non-negative constraints) .......... $\left(iv\right)$

The shaded region $OABC$ determined by linear inequalities $\left(ii\right)$ and $\left(iv\right)$ shown in the figure

Let us evaluate the objective function $Z$ at each point as shown below

At $O(0,0),$ $Z=8000\left(0\right)+12000\left(0\right)=0$

At $A(20,0),Z=8000\left(20\right)+12000\times 0=160000$\

At $B(12,6),Z=8000\times 20+12000\times 6=168000$

At $C(0,10),Z=8000\times 0+12000\times 10=120000$

Maximum value of $Z$ is $168000$ at $B(12,6)$

Hence, the company should produce $12$ pieces of Model $A$ and $6$ pieces of Model $B$ to get the maximum profit i.e. $168000$.

$\left(b\right)$

Given function

$f\left(x\right)=\begin{array}{c}\{\begin{array}{cc}Kx+1,& \text{if}x\ge 5\\ 3x-5,& \text{if}x\le 5\end{array}\end{array}$ is continuous at $x=5$

$⟹{lim}_{x\to 5^{-}}f\left(x\right)={lim}_{x\to 5^{+}}f\left(x\right)$

$⟹{lim}_{x\to 5^{-}}kx+1={lim}_{x\to 5^{+}}3x-5$

$⟹5k+1=15-5$

$⟹k=\frac{9}{5}$