IDNLearn.com is your go-to platform for finding reliable answers quickly. Our community is ready to provide in-depth answers and practical solutions to any questions you may have.
Sagot :
To solve the problem, we need to complete the table for different values of [tex]\( x \)[/tex] in the polynomial expression [tex]\( -7x^2 + 32x + 240 \)[/tex]. We will then determine the optimal price that the taco truck should sell its tacos for by identifying the value of [tex]\( x \)[/tex] that maximizes the polynomial expression. Here, [tex]\( x \)[/tex] represents the price in whole dollars.
### Step-by-Step Solution:
1. Understanding the Polynomial Expression:
The polynomial given is:
[tex]\[ f(x) = -7x^2 + 32x + 240 \][/tex]
2. Create the Table for Different Values of [tex]\( x \)[/tex]:
We'll evaluate the polynomial for [tex]\( x \)[/tex] ranging from 0 to 50.
- When [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -7(0^2) + 32(0) + 240 = 240 \][/tex]
- When [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = -7(1^2) + 32(1) + 240 = -7 + 32 + 240 = 265 \][/tex]
- When [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = -7(2^2) + 32(2) + 240 = -28 + 64 + 240 = 276 \][/tex]
- When [tex]\( x = 3 \)[/tex]:
[tex]\[ f(3) = -7(3^2) + 32(3) + 240 = -63 + 96 + 240 = 273 \][/tex]
- When [tex]\( x = 4 \)[/tex]:
[tex]\[ f(4) = -7(4^2) + 32(4) + 240 = -112 + 128 + 240 = 256 \][/tex]
- When [tex]\( x = 5 \)[/tex]:
[tex]\[ f(5)= -7(5^2) + 32(5) + 240 = -175 + 160 + 240 = 225 \][/tex]
- When [tex]\( x = 6 \)[/tex]:
[tex]\[ f(6) = -7(6^2) + 32(6) + 240 = -252 + 192 + 240 = 180 \][/tex]
Continuing this way up to [tex]\( x = 50 \)[/tex], we get the following table:
[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline 0 & 240 \\ 1 & 265 \\ 2 & 276 \\ 3 & 273 \\ 4 & 256 \\ 5 & 225 \\ 6 & 180 \\ 7 & 121 \\ 8 & 48 \\ 9 & -39 \\ 10 & -140 \\ 11 & -255 \\ 12 & -384 \\ 13 & -527 \\ 14 & -684 \\ 15 & -855 \\ 16 & -1040 \\ 17 & -1239 \\ 18 & -1452 \\ 19 & -1679 \\ 20 & -1920 \\ 21 & -2175 \\ 22 & -2444 \\ 23 & -2727 \\ 24 & -3024 \\ 25 & -3335 \\ 26 & -3660 \\ 27 & -3999 \\ 28 & -4352 \\ 29 & -4719 \\ 30 & -5100 \\ 31 & -5495 \\ 32 & -5904 \\ 33 & -6327 \\ 34 & -6764 \\ 35 & -7215 \\ 36 & -7680 \\ 37 & -8159 \\ 38 & -8652 \\ 39 & -9159 \\ 40 & -9680 \\ 41 & -10215 \\ 42 & -10764 \\ 43 & -11327 \\ 44 & -11904 \\ 45 & -12495 \\ 46 & -13100 \\ 47 & -13719 \\ 48 & -14352 \\ 49 & -14999 \\ 50 & -15660 \\ \hline \end{array} \][/tex]
3. Determine the Optimal Price:
We observe that the value of [tex]\( f(x) \)[/tex] is maximized when [tex]\( x = 2 \)[/tex]. At this price point, [tex]\( f(x) \)[/tex] yields the highest value, which is [tex]\( 276 \)[/tex].
### Conclusion:
The optimal price that the taco truck should sell its tacos for is [tex]\( \boxed{2} \)[/tex] dollars. This price maximizes the polynomial expression, corresponding to a value of 276.
### Step-by-Step Solution:
1. Understanding the Polynomial Expression:
The polynomial given is:
[tex]\[ f(x) = -7x^2 + 32x + 240 \][/tex]
2. Create the Table for Different Values of [tex]\( x \)[/tex]:
We'll evaluate the polynomial for [tex]\( x \)[/tex] ranging from 0 to 50.
- When [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -7(0^2) + 32(0) + 240 = 240 \][/tex]
- When [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = -7(1^2) + 32(1) + 240 = -7 + 32 + 240 = 265 \][/tex]
- When [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = -7(2^2) + 32(2) + 240 = -28 + 64 + 240 = 276 \][/tex]
- When [tex]\( x = 3 \)[/tex]:
[tex]\[ f(3) = -7(3^2) + 32(3) + 240 = -63 + 96 + 240 = 273 \][/tex]
- When [tex]\( x = 4 \)[/tex]:
[tex]\[ f(4) = -7(4^2) + 32(4) + 240 = -112 + 128 + 240 = 256 \][/tex]
- When [tex]\( x = 5 \)[/tex]:
[tex]\[ f(5)= -7(5^2) + 32(5) + 240 = -175 + 160 + 240 = 225 \][/tex]
- When [tex]\( x = 6 \)[/tex]:
[tex]\[ f(6) = -7(6^2) + 32(6) + 240 = -252 + 192 + 240 = 180 \][/tex]
Continuing this way up to [tex]\( x = 50 \)[/tex], we get the following table:
[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline 0 & 240 \\ 1 & 265 \\ 2 & 276 \\ 3 & 273 \\ 4 & 256 \\ 5 & 225 \\ 6 & 180 \\ 7 & 121 \\ 8 & 48 \\ 9 & -39 \\ 10 & -140 \\ 11 & -255 \\ 12 & -384 \\ 13 & -527 \\ 14 & -684 \\ 15 & -855 \\ 16 & -1040 \\ 17 & -1239 \\ 18 & -1452 \\ 19 & -1679 \\ 20 & -1920 \\ 21 & -2175 \\ 22 & -2444 \\ 23 & -2727 \\ 24 & -3024 \\ 25 & -3335 \\ 26 & -3660 \\ 27 & -3999 \\ 28 & -4352 \\ 29 & -4719 \\ 30 & -5100 \\ 31 & -5495 \\ 32 & -5904 \\ 33 & -6327 \\ 34 & -6764 \\ 35 & -7215 \\ 36 & -7680 \\ 37 & -8159 \\ 38 & -8652 \\ 39 & -9159 \\ 40 & -9680 \\ 41 & -10215 \\ 42 & -10764 \\ 43 & -11327 \\ 44 & -11904 \\ 45 & -12495 \\ 46 & -13100 \\ 47 & -13719 \\ 48 & -14352 \\ 49 & -14999 \\ 50 & -15660 \\ \hline \end{array} \][/tex]
3. Determine the Optimal Price:
We observe that the value of [tex]\( f(x) \)[/tex] is maximized when [tex]\( x = 2 \)[/tex]. At this price point, [tex]\( f(x) \)[/tex] yields the highest value, which is [tex]\( 276 \)[/tex].
### Conclusion:
The optimal price that the taco truck should sell its tacos for is [tex]\( \boxed{2} \)[/tex] dollars. This price maximizes the polynomial expression, corresponding to a value of 276.
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Thank you for visiting IDNLearn.com. We’re here to provide clear and concise answers, so visit us again soon.
