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lakkis [162]
3 years ago
12

Ruhana owns a workshop where her team of technicians refurbishes tv sets and dvd players, and then she sells them for a profit.

She has set a weekly sales target of at least 35 tv sets or dvd players. Additionally, she must ensure that her team collectively works at least 100 hours each week. It takes 4 hours to refurbish a tv set and 2 hours to refurbish a dvd player.
it costs $75 to refurbish a tv set and $40 to refurbish a dvd player. If ruhana wants to minimize costs, which point represents the optimal number of tv sets and dvd players that her team should refurbish each week?

Mathematics
1 answer:
Gemiola [76]3 years ago
7 0

Answer: Ruhana and her team should refurbish 15 tv sets and 20 dvd players each week to minimize the cost.


Step-by-step explanation:

For the given situation let the number of tv sets be x and number of dvd players be y.

Now we have to minimize the cost as it costs $75 to refurbish a tv set and $40 to refurbish a dvd player.

i.e. Minimize z=75x+40y

it gives subject to the constraints

4x+2y≥100......(1)(100 hours each week. It takes 4 hours to refurbish a tv set and 2 hours to refurbish a dvd player.)

x+y≥35.......(2)(weekly sales target is to refurbish at least 35 tv sets or dvd players.)

To plot equation (1) we need to find coordinates of points lying on line (1)

put x=0 gives 2y=100⇒y=50

put y=0 gives 4x=100⇒x=25

So we got points (0,50) and(25,0) for (1)..............(3)

Similarly for equation (2)

put x=0 gives y=35

put y=0 gives x=35

So we got points (0,35) and(35,0) for (2).................(4)

with the help of (3) and (4) we plot the following graph (assume x≥0 and y≥0)

The unbounded feasible region determined by constraints gives the corner points as A(0,50),B(15,20)and C(35,0).

from  we get the value of z is minimum at point B (15,20) .

hence we got our optimal solution at B (15,20), where x is the number of tv sets and y is the number of dvd players  .

therefore Ruhana and her team should refurbish 15 tv sets and 20 dvd players each week to minimize the cost.

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