$A$ factory manufactures two types of screws, $A$ and $B$. Each type of screw requires the use of two machines, an automatic and a hand-operated one. It takes $4 \, \text{minutes}$ on the automatic and $6 \, \text{minutes}$ on the hand-operated machine to manufacture a package of screws $A$, while it takes $6 \, \text{minutes}$ on the automatic and $3 \, \text{minutes}$ on the hand-operated machine to manufacture a package of screws $B$. Each machine is available for at most $4 \, \text{hours}$ on any day. The manufacturer can sell a package of screws $A$ at a profit of $Rs. \, 7$ and screws $B$ at a profit of $Rs. \, 10$. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximize his profit? Determine the maximum profit.
- A$Rs. \, 410$
- B$Rs. \, 400$
- C$Rs. \, 280$
- D$Rs. \, 350$
Explore More
Similar Questions
$A$ manufacturer produces two models of bikes: Model $X$ and Model $Y$. Model $X$ takes $6$ man-hours to make per unit,while Model $Y$ takes $10$ man-hours per unit. There is a total of $450$ man-hours available per week. Handling and marketing costs are $Rs. 2000$ and $Rs. 1000$ per unit for Models $X$ and $Y$ respectively. The total funds available for these purposes are $Rs. 80,000$ per week. Profits per unit for Models $X$ and $Y$ are $Rs. 1000$ and $Rs. 500$ respectively. How many bikes of each model should the manufacturer produce so as to yield a maximum profit? Find the maximum profit.
Difficult
View Solution$A$ manufacturing company produces two items,$A$ and $B$. Each item must be processed by two machines,$I$ and $II$. Machine $I$ can be operated for a maximum of $10$ hours $40$ minutes ($640$ minutes). It takes $20$ minutes for an item $A$ and $15$ minutes for an item $B$. Machine $II$ can be operated for a maximum of $8$ hours $20$ minutes ($500$ minutes). It takes $5$ minutes for an item $A$ and $8$ minutes for an item $B$. The profit per item of $A$ is ₹ $25$ and per item of $B$ is ₹ $18$. The formulation of an $L.P.P.$ to maximize the profit (where $x$ is the number of items $A$ and $y$ is the number of items $B$) is . . . . . . .
The graph with the correct feasible region of the $L.P.P.$ for the constraints $2x + y \leqslant 10$,$y \leqslant x$,$y \leqslant 2$,$x, y \geqslant 0$ is $\ldots$
If $Z = 7x + y$ subject to $5x + y \geq 5$,$x + y \geq 3$,$x \geq 0$,$y \geq 0$,then the minimum value of $Z$ is
The minimum value of $z = 10x + 25y$ subject to the constraints $0 \leq x \leq 3$,$0 \leq y \leq 3$,and $x + y \geq 5$ is:
DifficultMHT CET 2019
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo