Let's determine the daily revenue for each value of [tex]\( x \)[/tex] using the polynomial expression [tex]\( -7x^2 + 32x + 240 \)[/tex].
To find the daily revenue, substitute each value of [tex]\( x \)[/tex] (representing the taco price in dollars) into the polynomial [tex]\( -7x^2 + 32x + 240 \)[/tex] and calculate the resulting revenue.
Here's the table with calculations:
[tex]\[
\begin{array}{|c|c|c|}
\hline \text{Value of } x & \text{Taco Price (\$)} & \text{Daily Revenue (\$)} \\
\hline 0 & 0 & 240 \\
\hline 1 & 1 & 265 \\
\hline 2 & 2 & 276 \\
\hline 3 & 3 & 273 \\
\hline 4 & 4 & 256 \\
\hline 5 & 5 & 225 \\
\hline 6 & 6 & 180 \\
\hline
\end{array}
\][/tex]
Now, we seek to determine the optimal price that maximizes the daily revenue. From the calculated revenues, you can observe:
- At [tex]\( x = 0 \)[/tex], the revenue is \[tex]$240.
- At \( x = 1 \), the revenue is \$[/tex]265.
- At [tex]\( x = 2 \)[/tex], the revenue is \[tex]$276.
- At \( x = 3 \), the revenue is \$[/tex]273.
- At [tex]\( x = 4 \)[/tex], the revenue is \[tex]$256.
- At \( x = 5 \), the revenue is \$[/tex]225.
- At [tex]\( x = 6 \)[/tex], the revenue is \[tex]$180.
The maximum revenue is \$[/tex]276, which occurs when the taco price [tex]\( x \)[/tex] is \[tex]$2.
Hence, the optimal price that the taco truck should sell its tacos for is \$[/tex]2.
