Correct Option: CDetailed Step-by-Step Solution:The given expression is $$(4x+3y)\left(\frac{1}{x}+\frac{1}{y}\right)$$Step 1: Expand the expression.$$(4x+3y)\left(\frac{1}{x}+\frac{1}{y}\right) = 4x\left(\frac{1}{x}\right) + 4x\left(\frac{1}{y}\right) + 3y\left(\frac{1}{x}\right) + 3y\left(\frac{1}{y}\right)$$$$= 4 + \frac{4x}{y} + \frac{3y}{x} + 3$$$$= 7 + \frac{4x}{y} + \frac{3y}{x}$$Step 2: Apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality to the variable terms.For any two positive real numbers $a$ and $b$, the AM-GM inequality states that $$\frac{a+b}{2} \ge \sqrt{ab}$$which implies $$a+b \ge 2\sqrt{ab}$$.Let $a = \frac{4x}{y}$ and $b = \frac{3y}{x}$. Since $x$ and $y$ are positive real numbers, $a$ and $b$ are also positive.Applying the AM-GM inequality to these terms:$$\frac{4x}{y} + \frac{3y}{x} \ge 2\sqrt{\left(\frac{4x}{y}\right)\left(\frac{3y}{x}\right)}$$$$\ge 2\sqrt{\frac{4x \cdot 3y}{y \cdot x}}$$$$\ge 2\sqrt{12}$$$$\ge 2\sqrt{4 \cdot 3}$$$$\ge 2 \cdot 2\sqrt{3}$$$$\ge 4\sqrt{3}$$Step 3: Combine with the constant term to find the minimum value of the expression.The minimum value of $\left(\frac{4x}{y} + \frac{3y}{x}\right)$ is $4\sqrt{3}$.Therefore, the minimum value of the entire expression is $$7 + 4\sqrt{3}$$Step 4: Determine the condition for equality.The equality in AM-GM holds when $a=b$, i.e., when $$\frac{4x}{y} = \frac{3y}{x}$$$$4x^2 = 3y^2$$$$\frac{x^2}{y^2} = \frac{3}{4}$$$$\frac{x}{y} = \frac{\sqrt{3}}{2}$$ (since $x, y > 0$)Since positive real numbers $x$ and $y$ satisfying this condition exist, the minimum value is attainable.Why other options are incorrect:A) $7+2\sqrt{3}$: This option incorrectly calculates the product under the square root or misapplies the AM-GM inequality.B) $12$: This value does not correspond to the minimum derived from the AM-GM inequality. For instance, if one mistakenly assumes a product of terms yields an integer, this could be a distracter.D) $14$: This value is obtained if we assume $x=y$. In that scenario, the expression becomes $$(4x+3x)\left(\frac{1}{x}+\frac{1}{x}\right) = (7x)\left(\frac{2}{x}\right) = 14$$. However, $x=y$ is not the condition for the minimum value of the variable part $\frac{4x}{y} + \frac{3y}{x}$, as the minimum occurs when $\frac{x}{y} = \frac{\sqrt{3}}{2} \ne 1$.