Justin owns a food truck that sells tacos and burritos. He only has enough supplies to make 110 tacos or burritos. He sells each taco for $3.25 and each burrito for $5.75. Justin must sell a minimum of $520 worth of tacos and burritos each day. If 39 tacos were sold, determine the maximum number of burritos that the Justin could sell that would meet the requirements. If there are no possible solutions, submit an empty answer.

so looks its an easy one

110 - 39 = 71

71 x 5.75 = 408.25

39 x 3.25 = 126.75

408.25 +126.75 = 535

71 burritos and 39 tacos = 535 dollars

71

Step-by-step explanation:

\underline{\text{Define Variables:}}

Define Variables:

May choose any letters.

\text{Let }t=

Let t=

\,\,\text{the number of tacos sold}

the number of tacos sold

\text{Let }b=

Let b=

\,\,\text{the number of burritos sold}

the number of burritos sold

\text{\textquotedblleft can make 110 tacos/burritos"}\rightarrow \text{110 or fewer tacos/burritos}

“can make 110 tacos/burritos"→110 or fewer tacos/burritos

Use a \le≤ symbol

Therefore the total number of tacos and burritos sold, t+bt+b, must be less than or equal to 110:110:

t+b\le 110

t+b≤110

\text{\textquotedblleft a minimum of \$520"}\rightarrow \text{\$520 or more}

“a minimum of $520"→$520 or more

Use a \ge≥ symbol

Each taco sells for $3.25, so tt tacos will bring in 3.25t3.25t dollars. Each burrito sells for $5.75, so bb burritos will bring in 5.75b5.75b dollars. Therefore, the total amount of revenue 3.25t+5.75b3.25t+5.75b must be greater than or equal to \$520:$520:

3.25t+5.75b\ge 520

3.25t+5.75b≥520

\text{Plug in }\color{green}{39}\text{ for }t\text{ and solve each inequality:}

Plug in 39 for t and solve each inequality:

Justin sold 39 tacos

\begin{aligned}t+b\le 110\hspace{10px}\text{and}\hspace{10px}&3.25t+5.75b\ge 520 \\ \color{green}{39}+b\le 110\hspace{10px}\text{and}\hspace{10px}&3.25\left(\color{green}{39}\right)+5.75b\ge 520 \\ b\le 71\hspace{10px}\text{and}\hspace{10px}&126.75+5.75b\ge 520 \\ \hspace{10px}&5.75b\ge 393.25 \\ \hspace{10px}&b\ge 68.39 \\ \end{aligned}

t+b≤110and

39+b≤110and

b≤71and

3.25t+5.75b≥520

3.25(39)+5.75b≥520

126.75+5.75b≥520

5.75b≥393.25

b≥68.39

\text{The values of }b\text{ that make BOTH inequalities true are:}

The values of b that make BOTH inequalities true are:

\{69,\ 70,\ 71\}

{69, 70, 71}

Therefore the maximum number of burritos that Justin must sell is 71.

a 10 percent increase

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step-by-step explanation:

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