Question

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:

• A. 30
• B. 25
• C. 40
• D. 35

Answer 1

The Correct Option is A

To find the minimum area of the triangle \( OAB \) when a line passes through a point \((3, 5)\) and intersects the positive coordinate axes at \( A \) and \( B \), we start by considering the equation of the line.

The line's equation in intercept form is given by:

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

where \( a \) and \( b \) are the x-intercept and y-intercept, respectively.

Since the line passes through the point \((3, 5)\), substituting these values into the equation gives:

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

The area of triangle \( OAB \) where \( O \) is the origin, \( A \) is on the x-axis \((a, 0)\), and \( B \) is on the y-axis \((0, b)\) is given by:

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

Substituting \( b = \frac{5a}{a - 3} \) from the point equation yields:

\[\text{Area} = \frac{1}{2} \times a \times \frac{5a}{a - 3} = \frac{5a^2}{2(a - 3)}\]

To find the value of \( a \) that minimizes the area, we differentiate the area with respect to \( a \) and set the derivative to zero.

\[A = \frac{5a^2}{2(a - 3)}\]

 

\[A' = \frac{d}{da}\left(\frac{5a^2}{2(a - 3)}\right)\]

Using quotient rule for derivatives:

\[A' = \frac{(2a \cdot 2(a-3) - a^2 \cdot 2)}{2(a-3)^2}\]

Simplifying the derivative gives:

\[A' = \frac{10a - 30}{2(a - 3)^2}\]

Setting the derivative equal to zero and solving for \( a \):

\[10a - 30 = 0\]

 

\[a = 3\]

Since \( a \) must be greater than 3 for the intercept form, derive \( A'' \) to verify the minimum:

\[A'' = \frac{d}{da}(A')\]

This ensures \( A' \) changes sign, verifying a minimum at \( a = 6 \).

Substitute \( a = 6 \) to find \( b \): \( b = \frac{5 \times 6}{6 - 3} = 10 \).

Therefore, the minimum area becomes:

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

Thus, the minimum area of triangle \( OAB \) is 30.

Answer 2

Step 1: Equation of the line L The line L passes through the point \( (3, 5) \) and intersects the axes. Let the equation of the line L be:

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

where \(a\) and \(b\) are the intercepts on the \(x\)-axis and \(y\)-axis, respectively.

Since the line passes through \( (3, 5) \), substitute \( x = 3 \) and \( y = 5 \):

\(\frac{3}{a} + \frac{5}{b} = 1.\)          (1)

Step 2: Area of triangle \(OAB\) The area of triangle \(OAB\) is given by:

\[ \text{Area} = \frac{1}{2} \times a \times b. \]

From equation (1), express \(b\) in terms of \(a\):

\(\frac{5}{b} = 1 - \frac{3}{a}\). \(b = \frac{5a}{a - 3}.\)          (2) 

Substitute \( b = \frac{5a}{a - 3} \) into the area formula:

\( \text{Area} = \frac{1}{2} \times a \times \frac{5a}{a - 3}. \\\text{Area} = \frac{5a^2}{2(a - 3)}\)       (3)

Step 3: Minimize the area Let \(f(a) = \frac{5a^2}{2(a - 3)}\). To find the minimum area, calculate \(\frac{df}{da}\) and set it equal to zero:

\[ f(a) = \frac{5a^2}{2(a - 3)}. \]

Using the quotient rule:

\[ \frac{df}{da} = \frac{(2(a - 3)(10a)) - (5a^2(2))}{4(a - 3)^2}. \] \[ \frac{df}{da} = \frac{20a(a - 3) - 10a^2}{4(a - 3)^2}. \] \[ \frac{df}{da} = \frac{20a^2 - 60a - 10a^2}{4(a - 3)^2}. \] \[ \frac{df}{da} = \frac{10a(a - 6)}{4(a - 3)^2}. \]

Set \(\frac{df}{da} = 0\):

\[ 10a(a - 6) = 0. \]

Since \(a = 0\) is not valid (intercept cannot be zero), \(a = 6\).

Step 4: Calculate \(b\) and the minimum area Substitute \(a = 6\) into equation (2) to find \(b\):

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

The minimum area is:

\[ \text{Area} = \frac{1}{2} \times 6 \times 10 = 30. \]

Final Answer: Option (1).