sat-dog

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 $x=2y+11$. If the product of the two integers is $546$, then $xy=546$. Substituting $2y+11$ for $x$ in this equation results in $\left(2y+11\right)y=546$. Distributing the $y$ to both terms in the parentheses results in $2y^2+11y=546$. Subtracting $546$ from both sides of this equation results in $2y^2+11y-546=0$. The left-hand side of this equation can be factored by finding two values whose product is $2\left(-546\right)$, or $-\mathrm{1,092}$, and whose sum is $11$. The two values whose product is $-\mathrm{1,092}$ and whose sum is $11$ are $39$ and $-28$. Thus, the equation $2y^2+11y-546=0$ can be rewritten as $2y^2+28y-39y-546=0$, which is equivalent to $2y\left(y-14\right)+39\left(y-14\right)=0$, or $\left(2y+39\right)\left(y-14\right)=0$. By the zero product property, it follows that $2y+39=0$ and $y-14=0$. Subtracting $39$ from both sides of the equation $2y+39=0$ yields $2y=-39$. 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 $y=14$. Substituting $14$ for $y$ in the equation $xy=546$ yields $x\left(14\right)=546$. Dividing both sides of this equation by $14$ results in $x=39$. 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.