Find the minimum and maximum values of the function $\newline$ $y=26t-t^{2}$ on $[14,2]$ by comparing values at the critical points and endpoints. $\newline$ (Use symbolic notation and fractions where needed. If the function does not have extreme values, enter DNE.)

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Q. Find the minimum and maximum values of the function

\newline

y=26t-t^{2}

[14,2]

by comparing values at the critical points and endpoints.

\newline

(Use symbolic notation and fractions where needed. If the function does not have extreme values, enter DNE.)

Find Critical Points: To find the extreme values of the function $y = 26t - t^2$ on the interval $[14, 26]$ , we first need to find its critical points by taking the derivative and setting it equal to zero.
Derivative Calculation: The derivative of $y$ with respect to $t$ is $\frac{dy}{dt} = 26 - 2t$ . We set this equal to zero to find the critical points: $26 - 2t = 0$ .
Evaluate Endpoints: Solving for $t$ gives us $t = \frac{26}{2}$ , which simplifies to $t = 13$ . However, since $13$ is not in the interval $[14, 26]$ , it is not a critical point we need to consider for this problem.
Maximum Value Comparison: Since there are no critical points within the interval $[14, 26]$ , we only need to compare the values of the function at the endpoints of the interval to find the extreme values.
Minimum Value Comparison: We evaluate the function at the lower endpoint, $t = 14$ : $y(14) = 26(14) - (14)^2 = 364 - 196 = 168$ .
Minimum Value Comparison: We evaluate the function at the lower endpoint, $t = 14$ : $y(14) = 26(14) - (14)^2 = 364 - 196 = 168$ .We evaluate the function at the upper endpoint, $t = 26$ : $y(26) = 26(26) - (26)^2 = 676 - 676 = 0$ .
Minimum Value Comparison: We evaluate the function at the lower endpoint, $t = 14$ : $y(14) = 26(14) - (14)^2 = 364 - 196 = 168$ .We evaluate the function at the upper endpoint, $t = 26$ : $y(26) = 26(26) - (26)^2 = 676 - 676 = 0$ .Comparing the values at the endpoints, we find that the maximum value of the function on the interval $[14, 26]$ is $168$ at $t = 14$ , and the minimum value is $0$ at $t = 26$ .