# The Next Best Food Shack

### King of tacos

## What Is "The Next Best Food Shack"

## Tacos!!!!

It takes 2 minutes to make one meat taco and 5 minutes to make one fish taco. Ryan has 10 people on staff. One meat taco needs 2 people, while one fish taco needs 3. Ryan has the staff working for 8 hours. One meat taco(x) sells for $1.25 and fish taco(y) sells for $1.50. Everyday Ryan asks for at least 50 meat tacos. How many tacos should be sold to maximize profits.

x is the amount of meat tacos I should make.

y is the amount of fish tacos I should make.

x__>__50 y__>__0 (Tacos made per day)

These two small equations the current amount of tacos that are required of the staff to make. As you can see the meat tacos require at least 50 everyday. However, zero fish tacos are requested.

1.5x+1.25y=Profit (Money)

This equation represents the amount of money that each type of taco is sold for, and the profit that could be made by the minimum or the maximum amount of tacos sold.

2x+5y__<__480 (Time)

The meat tacos take 2 minutes to make. The fish tacos take 5 minutes to make. This equation shows how many tacos can be made in the 480 minutes that the staff is given to make the tacos.

2x+3y__<__10 (people)

It takes 2 people to put together a meat taco. However, it takes 3 people to make a fish taco. This equation shows how many tacos can be made at one time with the given number of people in the staff.

In the graph below this explaination, I was able to graph the above equations to provide you with the best production of tacos to create the maximum profit. The black dot represents the area of the maximum profit and how many meat and fish tacos it takes to reach that maximum production. The shaded area representing feasible region of profit.

Maximum Profit

X__>__140

Y__>__0

140*1.25__>__$175 per day

The above equation shows the maximum daily profit. The highest profit is made by selling at least 140 tacos per day. If each taco is sold for $1.25 each, The highest profit (if only 140 tacos are made) is $175 everyday.

Substitution

1.25x+1.50y=175

x=140

175+1.50 y=175

-175 -175

1.5y=0

y=0

Elimination

1.25x+1.50y=175

y=0

1.25x=175

x=140

## Pizza!!

For The Next Best Food Shack in order to keep serving pizza I need to make at least $300 everyday. Every day I make at the most 200 slices of cheese and 150 slices of pepperoni. One slice of cheese cost $1.00 and one slice of pepperoni cost $1.50. For every hour I sell at least 10 slices of cheese and 5 slices of pepperoni. I work 8 hours a day. In Order to maximize the amount of money I need to make, how much do I need to sell a day?

Step 1

x=the amount of cheese pizza.

y= the amount of pepperoni pizza.

x__<__200 y__<__150

Cost 1.00x + 1.50y__>__$300

I know this looks like a bunch of math but it’s actually my start up for business. It explains that I have no more than

200 slices of cheese and 150 slices of pepperoni for each day. It also shows how my cost represents how much pizza I sell for each day. For each Day I must make a least $300.

Step 2

x__<__200 y__<__150

y__>__.67x+200

My math explains on how I have created my final constraint for my graph. The equations show my where I start things on my graph.

Step 3

The Graph that was unfortunately unable to attach to this flyer represented all the points of intersections to see how I can maximize my product.

The points of intersection are ( 200,0)( 0,150) (0,200) (150,100) (150,200) ( 100, 200) ( 0, 300) (0,0)

My optimized equation is 150x+ 100y__>__$300 My equation basically says that in order to make $300 dollars I need to sell at least 150 slices of cheese and 100 slices of pepperoni.

So I took each point of intersection and multiplied it by the cost of either cheese or pepperoni pizza. I found out in order to maximize my product I need to sell 150 slices of cheese and 100 slices pepperoni pizzas a day. Even though I could sell more slices of people for a higher number, I don’t have the resources to create more pizza at the time. At the moment this is how I can maximize my product.

Substitution and Elimination. In order to find some points on my graph I needed to use substitution and elimination for solving equations.

Substitution

1.00x+1.50y=300

300. x=200

200 + 1.50 y=300

-200 -200

1.5y=100

y=150

Elimination

1.00x+1.50y=300. y=100

1.00x=150

x=150