Search any question & find its solution
Question:
Answered & Verified by Expert
A merchant plans to sell two types of personal computers-a desktop model and a portable model that will cost $₹ 25000$ and $₹ 40000$ respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchantshould stock to get maximum profit if he does not want to invest more than $₹ 70$ lakhs and if his profit on the desktop model is ₹ 4500 and on portable model is ₹ 5000 .
Solution:
1572 Upvotes
Verified Answer
Let there be $\mathrm{x}$ desktop and $\mathrm{y}$ portable computers. Total monthly demand of computer does not exceed 250 .
$\Rightarrow x+y \leq 250$, cost of 1 desktop computer if $₹ 25000$ and 1 portable computer is $₹ 40000$
$\therefore \quad$ cost of $x$ desktop and y portable computer $=₹(25000 x+40000 y)$
Maximum investment $=₹ 70$ lakhs $=₹ 70,00,000$ $\Rightarrow 25000 \mathrm{x}+40000 \mathrm{y} \leq 7000000$
or $5 x+8 y \leq 1400$
profit on 1 desktop computer is $₹ 4500$ and on 1 portable is $₹ 5000$, total profit,
$Z=4500 x+5000 y$
objective function to maximise $Z=4500 x+5000 y$ and constraints are $x+y \leq 250,5 x+8 y \leq 1400 x, y \geq 0$
(i) The line $x+y=250$ passes through
$\mathrm{A}(250,0), \mathrm{B}(0,250)$ putting $\mathrm{x}=0$ $\mathrm{y}=0$ in $\mathrm{x}+\mathrm{y} \leq 250,0 \leq 25$, which is true.
$\Rightarrow x+y \leq 250$ lies on or below $A B$.
(ii) The line $5 x+8 y=1400$ passes through $C(280,0)$, D $(0,175)$
put $x=0, y=0$ in $5 x+8 y \leq 14000,0 \leq 1400$ which is true. $\Rightarrow 5 x+8 y \leq 1400$ lies on and below $C D$.
(iii) $\mathrm{x} \geq 0$, lies on and to the right to axis.
(iv) $y \geq 0$, lies on and above $x$-axis.
The shaded area OAPD represent the feasible region where $\mathrm{P}$ is the point of intersection of
At $\mathrm{A}(250,0), \mathrm{Z}=4500 \mathrm{x}+5000 \mathrm{y}=4500 \times 250+0=1125000$
At $\mathrm{P}(200,50), \mathrm{Z}=4500 \times 200+5000 \times 50$
$=900000+250000=1150000$
At D $(0,175), \mathrm{Z}=0+5000 \times 175=875000$
$\Rightarrow$ Maximum profit of $₹ 1150000$ is obtained when he stocks 200 desktop computer and 50 portable.
$\Rightarrow x+y \leq 250$, cost of 1 desktop computer if $₹ 25000$ and 1 portable computer is $₹ 40000$
$\therefore \quad$ cost of $x$ desktop and y portable computer $=₹(25000 x+40000 y)$
Maximum investment $=₹ 70$ lakhs $=₹ 70,00,000$ $\Rightarrow 25000 \mathrm{x}+40000 \mathrm{y} \leq 7000000$
or $5 x+8 y \leq 1400$
profit on 1 desktop computer is $₹ 4500$ and on 1 portable is $₹ 5000$, total profit,
$Z=4500 x+5000 y$
objective function to maximise $Z=4500 x+5000 y$ and constraints are $x+y \leq 250,5 x+8 y \leq 1400 x, y \geq 0$
(i) The line $x+y=250$ passes through
$\mathrm{A}(250,0), \mathrm{B}(0,250)$ putting $\mathrm{x}=0$ $\mathrm{y}=0$ in $\mathrm{x}+\mathrm{y} \leq 250,0 \leq 25$, which is true.
$\Rightarrow x+y \leq 250$ lies on or below $A B$.
(ii) The line $5 x+8 y=1400$ passes through $C(280,0)$, D $(0,175)$
put $x=0, y=0$ in $5 x+8 y \leq 14000,0 \leq 1400$ which is true. $\Rightarrow 5 x+8 y \leq 1400$ lies on and below $C D$.
(iii) $\mathrm{x} \geq 0$, lies on and to the right to axis.
(iv) $y \geq 0$, lies on and above $x$-axis.
The shaded area OAPD represent the feasible region where $\mathrm{P}$ is the point of intersection of
At $\mathrm{A}(250,0), \mathrm{Z}=4500 \mathrm{x}+5000 \mathrm{y}=4500 \times 250+0=1125000$
At $\mathrm{P}(200,50), \mathrm{Z}=4500 \times 200+5000 \times 50$
$=900000+250000=1150000$
At D $(0,175), \mathrm{Z}=0+5000 \times 175=875000$
$\Rightarrow$ Maximum profit of $₹ 1150000$ is obtained when he stocks 200 desktop computer and 50 portable.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.