Choice B is correct. It’s given that function $g$ is defined by $g (t) = f (t) + 3$ and that $f (t) = 55 t - 2 t^{2}$ . Substituting $55 t minus 2 t squared$ for $f (t)$ in $g (t) = f (t) + 3$ yields $g (t) = 55 t - 2 t^{2} + 3$ , or $g (t) = - 2 t^{2} + 55 t + 3$ . The maximum value of $g (t)$ can be found by completing the square to rewrite the equation defining $g$ in the form $g (t) = a {(t - h)}^{2} + k$ , where the maximum value of the function is $k$ , which occurs when $t = h$ , and $a$ is a negative constant. The equation $g (t) = - 2 t^{2} + 55 t + 3$ is equivalent to $g (t) = - 2 (t^{2} - \frac{55}{2} t) + 3$ , which can be rewritten as $g (t) = - 2 (t^{2} - \frac{55}{2} t + {(\frac{55}{4})}^{2}) + 3 + 2 {(\frac{55}{4})}^{2}$ , or $g (t) = - 2 {(t - \frac{55}{4})}^{2} + 3 + 2 {(\frac{55}{4})}^{2}$ . This equation is in the form $g (t) = a {(t - h)}^{2} + k$ , where $a = - 2$ , $h = \frac{55}{4}$ , and $k = 3 + 2 {(\frac{55}{4})}^{2}$ . Thus, the maximum value of $g (t)$ is $3 + {2 (\frac{55}{4})}^{2}$ .
Alternate approach: Since the function $f$ is a quadratic function, the maximum value of $f (t)$ occurs at the value of $t$ that’s halfway between the two zeros of the function. The zeros of function $f$ can be found by substituting $0$ for $f (t)$ in the equation defining $f$ , which yields $0 = 55 t - 2 t^{2}$ . This equation can be rewritten as $0 = t (55 - 2 t)$ . By the zero product property, it follows that $t equals 0$ or $55 - 2 t = 0$ . Subtracting $55$ from each side of the equation $55 - 2 t = 0$ yields $- 2 t = - 55$ . Dividing each side of this equation by $negative 2$ yields $t = \frac{55}{2}$ . Therefore, the zeros of function $f$ are $0$ and $StartFraction 55 Over 2 EndFraction$ . The value that’s halfway between $0$ and $StartFraction 55 Over 2 EndFraction$ can be found by calculating the average of $0$ and $StartFraction 55 Over 2 EndFraction$ , which is $\frac{0 + \frac{55}{2}}{2}$ , or $StartFraction 55 Over 4 EndFraction$ . It follows that the maximum of function $f$ occurs when $t = \frac{55}{4}$ . Substituting $StartFraction 55 Over 4 EndFraction$ for $t$ in the equation defining function $f$ yields $f (\frac{55}{4}) = 55 (\frac{55}{4}) - 2 {(\frac{55}{4})}^{2}$ , which is equivalent to $f (\frac{55}{4}) = \frac{55^{2}}{4} - 2 (\frac{55^{2}}{4^{2}})$ . Multiplying $StartFraction 55 squared Over 4 EndFraction$ by $\frac{4}{4}$ in this equation to get a common denominator yields $f (\frac{55}{4}) = 4 (\frac{55^{2}}{4^{2}}) - 2 (\frac{55^{2}}{4^{2}})$ , or $f (\frac{55}{4}) = 2 (\frac{55^{2}}{4^{2}})$ , which is equivalent to $f (\frac{55}{4}) = 2 {(\frac{55}{4})}^{2}$ . Thus, the maximum value of $f (t)$ is $2 left parenthesis StartFraction 55 Over 4 EndFraction right parenthesis squared$ . Since the equation defining $g (t)$ is $g (t) = f (t) + 3$ , the maximum value of $g (t)$ is $3$ greater than the maximum value of $f (t)$ . It follows that the maximum value of $g (t)$ is $3 + 2 {(\frac{55}{4})}^{2}$ .

Choice A is incorrect and may result from conceptual or calculation errors.

Choice C is incorrect and may result from conceptual or calculation errors.

Choice D is incorrect and may result from conceptual or calculation errors.