Question

David is running a fried chicken stand at fall music festivals. He sells fried chicken legs for $4 each and fried chicken tenders for $8/ cup. A festival costs $60 for a vendor license and supply costs are $1 for each chicken leg and $2 for each cup of tenders. David wants to make profit of more than $300 but he only has $110 to spend on costs ahead of time. Create a total profit and a cost equation to model the situation with x = # of chicken legs and y = # cups of tenders.

Answer 1

From the question,

Chicken legs cost $1, but the selling price is $4

Chicken tender cost $2 per cup, but the selling price is $8

Now, a festival costs $60 and David has only $110 to spend.

Also number of chicken legs sold is represented as x and

number of chicken tenders sold is represented as y.

Hence the cost equation becomes

`\begin{gathered} x*1\text{ dollar for chicken legs + y}*2\text{ dollars for chicken tender + 60 }\leq110 \\ x+2y+60\leq110 \end{gathered}`

Note that profit = sales - cost

So we have to subtract the cost from the sales.

Now, David wants to make sales more than $300.

Hence the sales equation becomes

`\begin{gathered} x*4\text{ dollars for chicken legs + y}*8\text{ }*\text{dollars for chicken tender }\ge300 \\ 4x+8y\ge300 \end{gathered}`

So, we will subtract the cost equation from the sales equation to get the profit equation. This becomes

`\begin{gathered} 4x+8y-(x+2y+60)\ge300 \\ 4x+8y-x-2y-60\ge300 \\ 4x-x+8y-2y\ge300+60 \\ 3x+6y\ge360 \end{gathered}`

Hence, the cost and profit equation is

`\begin{gathered} 60+x+2y\leq110 \\ 3x+6y\ge360 \end{gathered}`

But what we have as a correct choice in the answers is the cost and sales equation, which is

`\begin{gathered} 60+x+2y\leq110 \\ 4x+8y\ge300 \end{gathered}`