A variable line `L` passes through the point `(3, 5)` and intersects the positive coordinate axes at the points `A` and `B`. The minimum area of the triangle `OAB`, where `O` is the origin, is:

Options

A. 30
B. 40
C. 35
D. 25

Answer

To find the minimum area of triangle OAB formed by a variable line passing through the point (3, 5) and intersecting the positive coordinate axes at points A and B, we can follow these steps:

### Step 1: Equation of the line
The equation of the line in intercept form is given by:

\[ \frac{x}{a} + \frac{y}{b} = 1 \]

where \(a\) is the x-intercept and \(b\) is the y-intercept. Since the line passes through the point (3, 5), we can substitute \(x = 3\) and \(y = 5\) into the equation:

\[ \frac{3}{a} + \frac{5}{b} = 1 \]

### Step 2: Express one variable in terms of the other
From the equation \( \frac{3}{a} + \frac{5}{b} = 1 \), we can express \(b\) in terms of \(a\):

\[ 5 = b - \frac{5 \cdot 3}{a} \]

Rearranging gives:

\[ 3b + 5a = ab \]

This can be rearranged to express \(a\) in terms of \(b\):

\[ a = \frac{3b}{b - 5} \]

### Step 3: Area of triangle OAB
The area \(A\) of triangle OAB can be expressed as:

\[ A = \frac{1}{2} \times a \times b \]

Substituting \(a\) from the previous step:

\[ A = \frac{1}{2} \times \left(\frac{3b}{b - 5}\right) \times b = \frac{3b^2}{2(b - 5)} \]

### Step 4: Differentiate to find minimum area
To find the minimum area, we differentiate \(A\) with respect to \(b\) and set the derivative to zero:

Let \(A(b) = \frac{3b^2}{2(b - 5)}\).

Using the quotient rule, we differentiate:

\[ \frac{dA}{db} = \frac{(6b)(b - 5) - (3b^2)(1)}{2(b - 5)^2} \]

Setting the numerator equal to zero:

\[ 6b(b - 5) - 3b^2 = 0 \]

This simplifies to:

\[ 3b^2 - 30b = 0 \]

Factoring gives:

\[ 3b(b - 10) = 0 \]

Thus, \(b = 0\) or \(b = 10\). Since \(b\) must be positive, we take \(b = 10\).

### Step 5: Find corresponding \(a\)
Now we substitute \(b = 10\) back into the equation for \(a\):

\[ a = \frac{3(10)}{10 - 5} = \frac{30}{5} = 6 \]

### Step 6: Calculate the minimum area
Now we can calculate the minimum area:

\[ A = \frac{1}{2} \times 6 \times 10 = 30 \]

Thus, the minimum area of triangle OAB is:

\[ \boxed{30} \]