Let the pieces of type A manufactured per week be x

Let the pieces of type B manufactured per week be y

$ \therefore $ maximum profit Z = $80 x + 120 y$

Fabricating hours for A is 9 and finishing hours is 1

Fabricating hours for B is 12 and finishing hours is 3

Maximum number of fabricating hours = 180

$ \therefore 9x + 12y \leq 180$

i.e., $ 3x+4y \leq 60$

Maximum number of finishing hours = 30

$ \therefore x + 3y \leq 30$

Now Z is $ = 80x+120y$ is subjected to

Constraints $ 3x+4y \leq 60$

$ x + 3y \leq 30$

$x \geq 0$

$y \geq 0$

Now the area of the fesible region is as shown in the figure

$3x+4y=60 \quad \quad \therefore y = 6$

$(3)x+3y=30 \quad \quad and \: x = 12$

$ 3x+4y=60$

$ 3x+9y=30$

We can see that the points on the bounded region are

A(0, 15), B( 0, 10), C(20, 0) and D(12, 6)

Now let us calculate the maximum profit

$Points (x,y) \quad \quad \quad Z = 80x + 120y$

$A(0, 15) \quad \quad \quad \quad \quad Z = 80(0) + 120(15) = 1800$

$B(0, 10) \quad \quad \quad \quad \quad Z = 80(0) + 120(10) = 1200$

$C(20, 0) \quad \quad \quad \quad \quad Z = 80(20) + 120(0) = 1600$

$D(12, 6) \quad \quad \quad \quad \quad Z = 80(12) + 120(6) = 960+720 = 1680$

Hence the maximum profit is at (0, 15)

$ \therefore $ teaching aid A = 0 and teaching aid B = 15 has to be made.