SOLUTION: Christopher is making necklace and key chains. He can make a necklace in 0.5 hours and key chain in 0.25 hours and he has no more than 20 hours. The cost in making a necklace is $2

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Question 1068942: Christopher is making necklace and key chains. He can make a necklace in 0.5 hours and key chain in 0.25 hours and he has no more than 20 hours. The cost in making a necklace is $2 while for key chain it is $3.He could invest up-to $120.write a system of four linear inequalities andfind how many of each he could make? If he sells each necklace for $10 and keychain for $8, tell when he will get maximum revenue and maximum profit?
Answer by ikleyn(52794)   (Show Source): You can put this solution on YOUR website!
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Christopher is making necklace and key chains.
He can make a necklace in 0.5 hours and key chain in 0.25 hours and he has no more than 20 hours.
The cost in making a necklace is $2 while for key chain it is $3. He could invest up-to $120.
Write a system of four linear inequalities and find how many of each he could make?
If he sells each necklace for $10 and key chain for $8, tell when he will get maximum revenue and maximum profit?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let X be the number of necklaces and Y be the number of key chains.

Then the inequalities are

X >= 0;

Y >= 0;

0.5X + 0.25Y <= 20   (the restriction by time);

2X + 3Y <= 120       (the restriction by investing)

The inequalities describe and represent this quadrilateral in the first quadrant,
restricted by the x- and y-axes and the red and the green lines:





Plots y =  (red)  and y =  (green)


The vertices of this quadrilateral are 

P1 = (0,0)     = the origin of the coordinate system,

P2 = (0,60)    = the y-intersect of the green line,

P3 = (40,0)    = the x-intersect of the red   line,  and

P4 = (30,20)   = the intersection point of the red and the green lines.

                 (this intersection point is the solution of the system  0.5X + 0.25Y = 20  and  2X + 3Y = 120).

The solution for the maximum profit

The profit function is P(X,Y) = ($10 - $2)*X + ($8 - $3)Y = 8X + 5Y dollars.    (1)

To find the maximum profit, you should check the values of the profit function P(x,Y) at all four vertices and then compare four values.


The values are: 

at P1:  P(X,Y) = P(0,0)   = 8*0  + 5*0   =   0;

at P2:  P(X,Y) = P(0,60)  = 8*0  + 5*60  = 300;

at P3:  P(X,Y) = P(40,0)  = 8*40 + 5*0   = 320;

at P4:  P(X,Y) = P(30,20) = 8*30 + 5*20  = 340.


The maximal value of the profit function is at P4, and it gives the optimal solution: 

      The maximum profit is $340 and it is achieved when Christoper produces 30 necklaces and 20 key chains.

The solution for the maximum revenue

The revenue function is R(X,Y) = $10*X + $8*Y = 10X + 8Y dollars.               (2)

To find the maximum revenue, you should check the values of the revenue function R(x,Y) (2) at all four vertices and then compare four values.


The values are: 

at P1:  R(X,Y) = R(0,0)   = 10*0  + 8*0   =   0;

at P2:  R(X,Y) = R(0,60)  = 10*0  + 8*60  = 480;

at P3:  R(X,Y) = R(40,0)  = 10*40 + 8*0   = 400;

at P4:  R(X,Y) = R(30,20) = 10*30 + 8*20  = 460.


The maximal value of the revenue function is at P2, and it gives the optimal solution for the revenue: 

      The maximum revenue is $480 and it is achieved when Christoper produces 0 necklaces and 60 key chains.

As you see, maximum profit and maximum revenue are achieved at different points and assume different strategies.


The method I solved this problem is called "the linear programming method".

It may have different / (other) names, too.



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