Really lost on this, Math is really my weakest point.

The Blivit Electronic Company manufactures bleeps and peeps. manufacturing a bleep requires 2 hours on machine A and 1 hour on Machine B. Manufactoring a peep requires 1 hour on machine A and 1 Hour on machine B. Machine A cannot be used more than 7 hours a day, and machine B cannot be used more than 5 hours a day. if the profit from a bleep is 5$ and the profit from a peep is $4, how many of each should be produced to maximize profit?

Im supposed to find a complete solution with an objective quantity, constraints, and vertices...ughh.
Would REALLY appreciate some help, me and math go together like fire and ice.

rather than trying to create complex formulae just use simple case studies to solve this:

maximum number of "bleeps" you can produce would be 3 using A for 6 hours and B for 3 hours giving 15$ profit and leaving A one hour and B two 2 hours idle
which allows to manufacture one "peep" adding 4$ to the profit, so in this scenario machine B would stay one hour idle and the total profit would be 19$

maximum number of "peeps" you can produce would be 5 both machine for 5 hours, giving 20$ profit and leaving machine A idle for 2 hours

now can you ensure to use these two hours of idle time to improve the profit?

yes, if you produce two "peeps" less you can produce two "bleeps"
this gives you three "peeps" with 12$ profit and two "bleeps" with 10$ profit, a total profit of 22$ and full usage of the machines

yeah i know how to solve it but the problem is he expects me to use constraints, objective quantity, and to graph using constaints...then finding vertices of feasible set..its just ughh so complicated im not good at math

Alright so little update
I tried doing this myself and this is what i have so far.
Constraints: x+y<5
2x+y<7
Vertices on graph: (1,0) (3.5,0) (2,3) and (0,5)
I have the whole graph portion set up but the tutor i spoke with told me to "plug those into the profit equation" and i got my answer and im done..what does he/she mean?

So here's your graph:
http://fooplot.com/plot/llwfbb6l2h

The optimal point occurs at one of the corners of the feasible region created by these two lines.
To solve the problem by hand, you must test each corner point to see which one generates the most profit.

Profit is represented as follows:
$5Bleeps + $4Peeps.
Or, in terms of x and y: 5x + 4y.

Plug in each set of ordered pairs ex. (1,2) and see which pair results in the highest value when it has been plugged in to the above function.

Thanks alot really helped, finished.