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Home  >>  CBSE XII  >>  Math  >>  Linear Programming
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A toy company manufactures two types of dolls, A and B. Market tests and available resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is at most half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of Rs 12 and Rs 16 per doll respectively on dolls A and B, how many of each should be produced weekly in order to maximise the profit?

 

 
 

 

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Toolbox:
  • First formulate the objective function and identify the constraints from the problem statement, To solve a Linear Programming problem graphically, first plot the constraints for the problem. This is done by plotting the boundary lines of the constraints and identifying the points that will satisfy all the constraints.
  • Let $R$ be the feasible region for a linear programming problem and let $Z=ax+by$ be the objective function.When $z$ has an optimum value (maximum or minimum),where variables $x$ and $y$ are subject to constraints described by linear inequalities,this optimum value must occur at a corner point of the feasible region.
  • If R is bounded then the objective function Z has both a maximum and minimum value on R and each of these occur at corner points of R
Step 1:
Let $x$ dolls of type $A$ and $y$ dolls of type $B$ be produced to have the maximum profit.
The company makes profit of Rs.12 and Rs.16 per doll respectively on doll A and doll B
$\Rightarrow Z=12x+16y$
The production level of $x+y$ should not exceed 1200
$x+y\leq 1200$
The production level of dolls of type A exceeds three times the production of dolls of type $B$ by at most 600
$\Rightarrow x-3y\leq 600$
Demand for dolls of type B is at most of that for dolls of type A.
$y\leq \large\frac{x}{2}$
Hence we can say the LPP is subject to maximize $Z=12x+16y$ and to constraints $x+y\leq 1200,x-3y\leq 600,y\leq \large\frac{x}{2}$$x,y\geq 0$
Step 2:
(i) The line $x+y=1200$ passes through $A(1200,0),B(0,1200)$
Put $x=0,y=0$ in $x+y\leq 1200$ we get $0\leq 1200$ which is true.
$\therefore x+y\leq 1200$ lies on and below AB
(ii) The line $x-3y=600$,passes through $C(600,0),D(0,-200)$
Put $x=0,y=0$ in $x-3y\leq 600$
Clearly $0 \leq 600$ is true.
Hence $x-3y\leq 600$ lies above the line $CD$
(iii) $y=\large\frac{x}{2}$ passes through (400,200) and (0,0)
Put $x=200,y=0$ in $y\leq \large\frac{x}{2}$
$0\leq \large\frac{200}{2}$ is also true.
$\Rightarrow (200,0) lies in the region (i.e)below OP
(iv) $x\geq 0$ lies on and to the right of $y$-axis.
(v) $y\geq 0$ lies on and above $x$-axis.
The shaded portion PQCO represents the feasible region.
Step 3:
The point $P$ is in the intersection of the lines $x+y=1200,y=\large\frac{x}{2}$$\Rightarrow 2y=x$
On solving these two equations we get,
$x=800$ and $y=400$
The point $Q$ is in the intersection of the lines $x+y=1200$ and $x-3y=600$
Let us solve these two equations to get the value of $x$ and $y$
$x+y=1200$
$x-3y=600$
_________________
$4y=600$
$y=150$
Hence $x=1050$
The coordinates of $Q$ are (1050,150).
The point $C$ is $(600,0)$
The objective function is $Z=12x+16y$
Step 4:
Let us now obtain the values of objective function as follows :
At the points $(x,y)$ the value of the objective function $Z=12x+16y$
At $P(800,400)$ the value of the objective function $Z=12\times 800+16\times 400=16,000$
At $Q(1050,150)$ the value of the objective function $Z=12\times 1050+16\times 150=15,000$
At $C(600,0)$ the value of the objective function $Z=12\times 600+16\times 0=7200$
At $O(0,0)$ the value of the objective function $Z=12\times 0+16\times 0=0$
It is clear that $Z$ is maximum at $P(800,400)$.The maximum value of $Z$ is Rs16,000 .Thus to maximize the profit 800 dolls of type A and 400 dolls of type B should be produced to get a maximum profit of Rs. 16,000
answered Aug 29, 2013 by sreemathi.v
edited Jan 30, 2014 by balaji.thirumalai
 

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