The value of `a(agt0)` for which the area bounded by the curves `y=(x)/(6)+(1)/(x^(2)),y=0,x=a, and x=2a` has the least value is ___.

Explanation

To find the value of \( a \) (where \( a > 0 \)) for which the area bounded by the curves \( y = \frac{x}{6} + \frac{1}{x^2} \), \( y = 0 \), \( x = a \), and \( x = 2a \) has the least value, we can follow these steps:

### Step 1: Set Up the Area Function
The area \( A \) between the curves can be expressed as:
\[ A = \int_{a}^{2a} \left( \frac{x}{6} + \frac{1}{x^2} \right) dx \]

### Step 2: Evaluate the Integral
We need to evaluate the integral:
\[ A = \int_{a}^{2a} \left( \frac{x}{6} + \frac{1}{x^2} \right) dx \]
This can be split into two parts:
\[ A = \int_{a}^{2a} \frac{x}{6} \, dx + \int_{a}^{2a} \frac{1}{x^2} \, dx \]

### Step 3: Compute Each Integral
1. For the first integral:
\[ \int \frac{x}{6} \, dx = \frac{x^2}{12} \]
Evaluating from \( a \) to \( 2a \):
\[ \left[ \frac{(2a)^2}{12} - \frac{a^2}{12} \right] = \frac{4a^2}{12} - \frac{a^2}{12} = \frac{3a^2}{12} = \frac{a^2}{4} \]

2. For the second integral:
\[ \int \frac{1}{x^2} \, dx = -\frac{1}{x} \]
Evaluating from \( a \) to \( 2a \):
\[ \left[ -\frac{1}{2a} + \frac{1}{a} \right] = -\frac{1}{2a} + \frac{2}{2a} = \frac{1}{2a} \]

### Step 4: Combine the Results
Now, combine both parts to get the total area:
\[ A = \frac{a^2}{4} + \frac{1}{2a} \]

### Step 5: Minimize the Area Function
To find the value of \( a \) that minimizes the area \( A \), we differentiate \( A \) with respect to \( a \):
\[ \frac{dA}{da} = \frac{d}{da} \left( \frac{a^2}{4} + \frac{1}{2a} \right) = \frac{a}{2} - \frac{1}{2a^2} \]
Setting the derivative equal to zero for minimization:
\[ \frac{a}{2} - \frac{1}{2a^2} = 0 \]
Multiplying through by \( 2a^2 \) to eliminate the fraction:
\[ a^3 - 1 = 0 \]
Thus, we find:
\[ a^3 = 1 \implies a = 1 \]

### Step 6: Conclusion
The value of \( a \) for which the area is minimized is:
\[ \boxed{1} \]