The value of a such that the area bounded by the curve `y=x^(2)+2ax+3a^(2)`, the coordinate axes and the line `x=1`. The coordinate axes and the line `x=1`. Attains its least value is equal to

Options

A. `(1)/(6)`
B. `-(1)/(6)`
C. `(1)/(3)`
D. `-(1)/(3)`

Answer

To find the value of \( a \) such that the area bounded by the curve \( y = x^2 + 2ax + 3a^2 \), the coordinate axes, and the line \( x = 1 \) attains its least value, we can follow these steps:

### Step 1: Set up the area integral
The area \( A \) under the curve from \( x = 0 \) to \( x = 1 \) is given by the integral:

\[ A = \int_0^1 (x^2 + 2ax + 3a^2) \, dx \]

### Step 2: Calculate the integral
We can compute the integral term by term:

1. The integral of \( x^2 \) from 0 to 1:
\[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1^3}{3} - 0 = \frac{1}{3} \]

2. The integral of \( 2ax \) from 0 to 1:
\[ \int_0^1 2ax \, dx = 2a \left[ \frac{x^2}{2} \right]_0^1 = 2a \cdot \frac{1^2}{2} - 0 = a \]

3. The integral of \( 3a^2 \) from 0 to 1:
\[ \int_0^1 3a^2 \, dx = 3a^2 \left[ x \right]_0^1 = 3a^2 (1 - 0) = 3a^2 \]

Combining these results, we have:

\[ A = \frac{1}{3} + a + 3a^2 \]

### Step 3: Find the minimum area
To find the value of \( a \) that minimizes the area \( A \), we take the derivative of \( A \) with respect to \( a \) and set it to zero:

\[ \frac{dA}{da} = 1 + 6a \]

Setting the derivative equal to zero:

\[ 1 + 6a = 0 \implies 6a = -1 \implies a = -\frac{1}{6} \]

### Step 4: Verify it's a minimum
To confirm that this value of \( a \) gives a minimum area, we can check the second derivative:

\[ \frac{d^2A}{da^2} = 6 \]

Since \( \frac{d^2A}{da^2} > 0 \), this indicates that \( A \) is indeed minimized at \( a = -\frac{1}{6} \).

### Final Answer
Thus, the value of \( a \) such that the area attains its least value is:

\[ \boxed{-\frac{1}{6}} \]