Question

If a line is moving between the coordinate axes such that the sum of the intercepts made by it on the coordinate axes is always 12, then the equation of that line which forms a triangle of maximum area with the coordinate axes is:

• A. \( 3x + y = 9 \)
• B. \( 5x + 7y = 35 \)
• C. \( x + y = 6 \)
• D. \( 5x + y = 10 \)

Hint:

When maximizing the area of a triangle formed by a line and the coordinate axes, use the constraint \( a + b = \text{constant} \) and maximize the product \( ab \).

Answer 1

The Correct Option is C

We are given that the sum of the intercepts made by the line on the coordinate axes is always 12. The general form of the equation of a line with intercepts \( a \) and \( b \) on the x-axis and y-axis, respectively, is: \[ \frac{x}{a} + \frac{y}{b} = 1 \] The sum of the intercepts is given by \( a + b = 12 \). The area of the triangle formed by the line and the coordinate axes is: \[ \text{Area} = \frac{1}{2} \times a \times b \] To maximize the area, we need to maximize the product \( ab \), subject to the constraint \( a + b = 12 \). Using the method of Lagrange multipliers or by substituting \( b = 12 - a \), we maximize \( ab \) to get the maximum area when \( a = b = 6 \). Thus, the equation of the line is \( x + y = 6 \). Therefore, the correct answer is option (3), \( x + y = 6 \).

Answer 2

Step 1: General equation of the line with intercepts
Let the intercepts on the x-axis and y-axis be \( a \) and \( b \) respectively.
Equation of the line:
\[ \frac{x}{a} + \frac{y}{b} = 1 \]

Step 2: Given condition on intercepts
\[ a + b = 12 \]

Step 3: Area of triangle formed with axes
Area of triangle formed by line and coordinate axes:
\[ A = \frac{1}{2} \times a \times b \]

Step 4: Express area in terms of one variable
From \( a + b = 12 \), we get:
\[ b = 12 - a \]
So area:
\[ A = \frac{1}{2} a (12 - a) = 6a - \frac{a^2}{2} \]

Step 5: Maximize the area
Take derivative w.r.t \( a \):
\[ \frac{dA}{da} = 6 - a \]
Set derivative to zero for maximum:
\[ 6 - a = 0 \implies a = 6 \]
Then:
\[ b = 12 - 6 = 6 \]

Step 6: Equation of the line with maximum area
Substitute \( a = 6 \), \( b = 6 \):
\[ \frac{x}{6} + \frac{y}{6} = 1 \implies x + y = 6 \]