Question

The area of the region bounded by the lines $\frac{x}{7\sqrt{3a}} + \frac{y}{b} = 4$, $x = 0$, and $y = 0$ is:

• A. $\frac{56}{\sqrt{3ab}}$
• B. $56a$
• C. $\frac{ab}{2}$
• D. $3ab$

Answer 1

Answer: (A)

$\frac{56}{\sqrt{3ab}}$

Answer 2

The equation of the line is:

$\frac{x}{7\sqrt{3a}} + \frac{y}{b} = 4.$

Rewriting:

$b \cdot x + 7\sqrt{3a} \cdot y = 28\sqrt{3ab}.$

To find the intercepts:

1. When $x = 0$, $y = 4b$.
2. When $y = 0$, $x = 28\sqrt{3a}$.

The lines $x = 0$ and $y = 0$, together with $\frac{x}{7\sqrt{3a}} + \frac{y}{b} = 4$, form a triangle with vertices:

$(0, 0), \quad (28\sqrt{3a}, 0), \quad (0, 4b).$

The area of the triangle is:

$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height},$

where the base is $28\sqrt{3a}$ and the height is $4b$:

$\text{Area} = \frac{1}{2} \times (28\sqrt{3a}) \times (4b).$

Simplify:

$\text{Area} = \frac{1}{2} \times 112\sqrt{3ab} = 56\sqrt{3ab}.$

Thus, the area of the region is:

$56\sqrt{3ab}.$