# Thread: Linear Programming application

1. ## Linear Programming application

My problem is with constraints.

Here is the problem in a nutshell

1000 guests in a fundraiser. $1 for juice$1.75 for Soda
Constraints:

each drink has to be served in an unused plastic cup of which there are 5,000

juice must acount for at least 30% of drinks served

each guest can only have 3 sodas

each juice must have 2 icecubes and each soda must have 4 ice cubes, there are 15,000 icecubes available.

goal is to maximize profit

I know the objective function is going to be 1x+1.75y=z
I think one constraint is going to be y less than/equal to 3 and after that i get lost. please help

2. Hello, Unusualtoe!

This certainly has a tricky set-up . . .

1000 guests at a fundraiser.
$1 for juice,$1.75 for soda
Let $x$ = number of servings of juice sold: . ${\color{blue}x \geq 0}$
Let $y$ = number of servings of soda sold: . ${\color{blue}y \geq 0}$

Each drink has to be served in an unused plastic cup of which there are 5,000
We have: . ${\color{blue}x + y \:\leq \:5000}$

Juice must account for at least 30% of drinks served
It says: .Juice is at least 30% of (Juice + Soda)

So we have: . $x \:\geq \:0.30(x+y)$
. . This becomes: . $x \:\geq\:0.3x + 0.3y\quad\Rightarrow\quad {\color{blue}7x - 3y \:\geq\:0}$

Each guest can only have 3 sodas
There are 1000 guests.
If each has 3 sodas, 3000 sodas will be sold: . ${\color{blue}y \:\leq\:3000}$

Each juice must have 2 ice cubes and each soda must have 4 ice cubes.
There are 15,000 ice cubes available.
There will be: . $2x + 4y$ ice cubes used.
. . $2x + 4y \:\leq \:15,000\quad\Rightarrow\quad {\color{blue}x + 2y \:\leq \:7500}$

Goal is to maximize profit
Objective function: . ${\color{red}P \:=\:x + 1.75y}$

Graph the region determined by the six constrants (in blue).
Test the vertices in the Profit function (in red).