Algebra 2 (Linear Programming) Help!

A bicycle builder makes two models. The basic model requires 2 hours of frame construction, 4 hours of assembly, and 1 hour of finishing. The deluxe model requires 3 hours of frame construction, 3 hours of assembly, and 2 hours of finishing. Each day 36 hours are available for frame construction, 40 hours for assembly, and 20 hours for finishing. If the profit on the basic model is $100 and on the deluxe model is $150, how many of each should be made to maximize the profit?

Let x = number of basic model, y = number of deluxe model

Then the constraining equations are:

2x + 3y <= 36
4x + 3y <= 40
x + 2y <= 20

with the profit function P = 100x + 150y

The maximum value of P occurs at a vertex of the quadrilateral region formed by the three constraining equations in the first quadrant. The four vertices are (0,0), (10,0), (4,8), (0,10).

P(0,0)) = 0
P(10,0) = 1000
P(4,8) = 400 + 1200 = 1600
P(0,10) = 1500

The maximum profit is $1600 when x = 4 and y = 8.
So the answer to this problem is 4 basic models and 8 deluxe models in order to maximize profit.

Let x= number of basics
Y = number of deluxe.

P= 100x +150y

Constraints:
x>=0, y>=0

framing: two hours for each basic plus 3 hours for each deluxe has to total up to 36 hours:
2x+3y<=36
Assembly: 4x+3y<=40
finishing: 1x+2y<=20
So now you have to graph the three lines in quadrant I to find the feasible region. Locate all vertices, Where they intersect and shade the innermost region.
plug the vertices into the profit function.

3y<=-2x+36; y<= -2/3 x +12
3y<=-4x+40; y< = -4/3 x+ 40/3
2y<= -x+20; y<= -1/2 x+ 10

(0,0)
(0,10)
(10,0)
(4,8)
--------
P= 100x + 150y
(0,0)-> 0
(0,10)-> 1500
(10,0)-> 1000
(4,8)-> 400+ 1200= 1600***
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maximum profit is at (4basic and 8 deluxe)

Hoping this helps!