The point,at which the maximum value of $10x + 6y$ subject to the constraints $x + y \leq 12$,$2x + y \leq 20$,$x \geq 0$,$y \geq 0$ occurs,is

  • A
    $(10, 0)$
  • B
    $(8, 4)$
  • C
    $(0, 12)$
  • D
    $(12, 0)$

Explore More

Similar Questions

The minimum value of $Z = 5x + 8y$ subject to the constraints $x + y \geq 5$,$0 \leq x \leq 4$,$y \geq 2$ is:

$A$ manufacturer makes two types of toys $A$ and $B$. Three machines are needed for this purpose and the time (in $minutes$) required for each toy on the machines is given below:
Types of ToysMachine-$I$Machine-$II$Machine-$III$
$A$$12$$18$$6$
$B$$6$$0$$9$

Each machine is available for a maximum of $6 \, hours$ $(360 \, minutes)$ per day. If the profit on each toy of type $A$ is $Rs. \, 7.50$ and that on each toy of type $B$ is $Rs. \, 5$,find the number of toys of each type that should be manufactured in a day to get maximum profit.

Difficult
View Solution

The Linear Programming Problem ($L$.$P$.$P$.) to minimize $z = 30x + 20y$ subject to the constraints $x + y \leqslant 8$,$x + 2y \geqslant 4$,$6x + 4y \geqslant 12$,$x \geqslant 0$,and $y \geqslant 0$ has:

$A$ diet is to contain at least $80$ units of vitamin $A$ and $100$ units of minerals. Two foods $F_{1}$ and $F_{2}$ are available. Food $F_{1}$ costs $Rs. 4$ per unit and food $F_{2}$ costs $Rs. 6$ per unit. One unit of food $F_{1}$ contains $3$ units of vitamin $A$ and $4$ units of minerals. One unit of food $F_{2}$ contains $6$ units of vitamin $A$ and $3$ units of minerals. Formulate this as a linear programming problem. Find the minimum cost for a diet that consists of a mixture of these two foods and also meets the minimal nutritional requirements.

Difficult
View Solution

$A$ cottage industry manufactures pedestal lamps and wooden shades,each requiring the use of a grinding/cutting machine and a sprayer. It takes $2 \text{ hours}$ on the grinding/cutting machine and $3 \text{ hours}$ on the sprayer to manufacture a pedestal lamp. It takes $1 \text{ hour}$ on the grinding/cutting machine and $2 \text{ hours}$ on the sprayer to manufacture a shade. On any day,the sprayer is available for at most $20 \text{ hours}$ and the grinding/cutting machine for at most $12 \text{ hours}$. The profit from the sale of a lamp is $Rs. 5$ and that from a shade is $Rs. 3$. Assuming that the manufacturer can sell all the lamps and shades that he produces,how should he schedule his daily production in order to maximise his profit?

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo