The decision variables:

Let

C = number of packs of Colonial to be produced per week,

I = number of packs of Italiano to be produced per week.

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The constraints:

there should at most 700 dozen packages produced in total,

So,

C +I <= 700(12)

C +I <= 8400 ------------(1)

the number of packs of Colonial lampshades should not exceed the number of packs of Italiano lampshades by more than 350.

So,

C -I <= 350 ---------------(2)

...the amount of plastic available(500kg)

....Colonial, 1 plastic kg per pack

....Italiano, 0.5 plastic kg per pack

so,

C(1) +I(0.5) <= 500

2C +I <= 1000 -----------------(3)

...and processing time (40 hrs)

...Colonial, 3 production minutes per pack

...Italiano, 4 production minutes per pack

I assume "production minutes" is same as "processing minutes" here, so,

C(3/60) +I(4/60) <= 40

3C +4I <= 2400 ----------------(4)

The non-negative constraints,

C >= 0 -------------(5)

I >= 0 -------------(6)

I don't know how to graph here so graph the 6 constraints on paper. Say, horizontal axis is for C, so vertical axis is for I.

Let me assume you know how to do that.

Let me assume also that you know how to get the intersection of two inequalities.

The inequality C +I <= 8400 ------------(1) is way out of the feasible region, rendering it useless. But the feasible region is in the correct side of the said inequality.

I suspect then that it should have been for "at most 700 packages produced in total", not for 700 dozens packages as posted.

The feasible region is a pentagon whose 5 corner points, in the order (C,I), are:

(0,0) ---------intersection of (5) and (6)

(0,600) -------intersection of (6) and (4)

(320,360) -----intersection of (3) and (4)

(450,100) -----intersection of (3) and (2)

(350,0) -------intersection of (2) and (5)

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Objective: to maximize profit.

Objective function:

P = C(8.00) +I(5.00)

P = 8C +5I ---------------(i)

Test that on all 5 vertices:

(0,0) ======> P = 0

(0,600) ====> P = 8*0 +5*600 = 3000

(320,360) ==> P = 8*320 +5*360 = 4360

(450,100) ==> P = 8*450 +5*100 = 4100

(350,0) ====> P = 8*350 +5*0 = 2800

Therefore, maximum profit ($4360 per week) is when 320 packs of Colonial lampshades and 360 packs of Italiano lampshades are produced per week. -------answer.