Choice B is correct. Let $x$ be the first integer and let $y$ be the second integer. If the first integer is $11$ greater than twice the second integer, then $x = 2 y + 11$ . If the product of the two integers is $546$ , then $x y equals 546$ . Substituting $2 y plus 11$ for $x$ in this equation results in $(2 y + 11) y = 546$ . Distributing the $y$ to both terms in the parentheses results in $2 y^{2} + 11 y = 546$ . Subtracting $546$ from both sides of this equation results in $2 y^{2} + 11 y - 546 = 0$ . The left-hand side of this equation can be factored by finding two values whose product is $2 left parenthesis negative 546 right parenthesis$ , or $negative 1,092$ , and whose sum is $11$ . The two values whose product is $negative 1,092$ and whose sum is $11$ are $39$ and $negative 28$ . Thus, the equation $2 y^{2} + 11 y - 546 = 0$ can be rewritten as $2 y^{2} + 28 y - 39 y - 546 = 0$ , which is equivalent to $2 y (y - 14) + 39 (y - 14) = 0$ , or $(2 y + 39) (y - 14) = 0$ . By the zero product property, it follows that $2 y + 39 = 0$ and $y - 14 = 0$ . Subtracting $39$ from both sides of the equation $2 y + 39 = 0$ yields $2 y = - 39$ . Dividing both sides of this equation by $2$ yields $y = - \frac{39}{2}$ . Since $y$ is a positive integer, the value of $y$ is not $- \frac{39}{2}$ . Adding $14$ to both sides of the equation $y - 14 = 0$ yields $y equals 14$ . Substituting $14$ for $y$ in the equation $x y equals 546$ yields $x left parenthesis 14 right parenthesis equals 546$ . Dividing both sides of this equation by $14$ results in $x equals 39$ . Therefore, the two integers are $14$ and $39$ , so the smaller of the two integers is $14$ .

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

Choice C is incorrect. This is the larger of the two integers.

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