Question 1207808: Sophia is organizing a movie night and wants to provide popcorn and soda to her guests. A bag of popcorn costs her $2 to buy, and a can of soda costs her $1. She only has room to store 24 cans of soda and 50 bags of popcorn at her house. She needs to have at least 60 items total to satisfy her guests. What is the least amount of money she can spend on popcorn and soda? ( This is a graphing question so if you read this do you think you could email me and I will send you the pictures of the questions o have 3 questions and if it’s necessary I am willing to pay for it.) Thank you i need the constraints,objetive function, Vertices of possible max./min, solutions, and of course the solution and graph ! Thank you so much
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
x = number of bags of popcorn
y = number of cans of soda
Since each bag of popcorn costs $2, she spends 2x dollars on popcorn alone.
Each can of soda costs $1, meaning she spends 1y = y dollars on soda alone.
2x+y = total amount spent for both items combined
This will set up the objective function C = 2x+y.
The goal is to make the output of this function the smallest possible, i.e. we want the smallest possible cost.
"She only has room to store 24 cans of soda and 50 bags of popcorn at her house" gives us these constraints  and 
"She needs to have at least 60 items total to satisfy her guests" gives this constraint 
Here are all of the constraints listed as a system of inequalities
The first two inequalities define a rectangle that is 50 units across and 24 units tall.
The lower left corner of this rectangle is at the origin (0,0).
The upper right corner is at (50,24).
We'll overlap this rectangle with
This is the region where we shade above the line x+y = 60.
Two points on this line are (0,60) and (60,0)
Here's what the final region looks like
Any point in the purple region, or on the boundary, will satisfy all three constraint inequalities.
Graph created with Desmos.
Desmos link here
https://www.desmos.com/calculator/9svzgkteyb
The triangular region has the vertices:
(50,10)
(50,24)
(36,24)
These vertices are determined by intersecting the various boundary lines.
For example, solve the system {x+y=60,y=24} to determine the vertex location (36,24).
The idea is to plug those coordinates into the objective function to figure out which x,y pair gives the smallest value of C.
Let's try the vertex (50,10)
C = 2x+y
C = 2*50+10
C = 110
Now try the vertex (50,24)
C = 2x+y
C = 2*50+24
C = 124
So far the first vertex leads to the lowest cost.
Now the last vertex
C = 2x+y
C = 2*36+24
C = 96
This is the smallest cost of the trio.
Therefore, the lowest cost ($96) occurs when Sophia buys x = 36 bags of popcorn and y = 24 cans of soda.
Answer by ikleyn(52817)  (Show Source):
You can put this solution on YOUR website! .
Sophia is organizing a movie night and wants to provide popcorn and soda to her guests.
A bag of popcorn costs her $2 to buy, and a can of soda costs her $1.
She only has room to store 24 cans of soda and 50 bags of popcorn at her house.
She needs to have at least 60 items total to satisfy her guests.
What is the least amount of money she can spend on popcorn and soda?
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I will solve it using different method, applying logical reasoning and common sense.
I will not make any plots to save my time.
Sophia wants to minimize her spending, still satisfying all constraints.
Let's apply the most economic strategy.
Sophia should take as much cheep items (sodas) as possible (24 cans),
and then complement it with an appropriate number of more expensive items (popcorn bags).
So, she needs 60-24 = 36 popcorn bags.
Her spending will be $2*36 + $1*24 = $96,
and, obviously, it is minimal spending.
Notice that this solution satisfies all the constraints.
Thus we solved the problem easy and elegantly using logical reasoning and common sense.
No plots are needed.
---------------
Outwardly, this problem is similar to Linear Programming problems,
where the geometric solution method is often used.
But concretely, this particular problem can be easily solved in MUCH SIMPLER WAY
using simple logic without any plots. That's why I call such tasks False Linear Programming problems.
All of your other today's problems belong to the same class/type,
and can be solved using logic, ONLY.
In my view, this simple logical solution is much more valuable solution method for the given problem
than to apply heavy artillery of the Linear Programming method blindly and without necessity.
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