Math, asked by roshanrs8559, 10 hours ago

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?

Answers

Answered by sushmashedage3
0

Answer:

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 She and her team refurbish 15 tv sets and 20 DVD players each week to minimize the cost.

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