If `g(x)=max(y^(2)-xy)(0le yle1)`, then the minimum value of g(x) (for real x) is

Options

A. `(1)/(4)`
B. `3-sqrt3`
C. `3+sqrt8`
D. `(1)/(2)`

Explanation

To find the minimum value of the function \( g(x) = \max_{0 \leq y \leq 1} (y^2 - xy) \), we will analyze the expression \( y^2 - xy \) for the given range of \( y \).

### Step 1: Analyze the expression
The expression we need to maximize is:
\[ f(y) = y^2 - xy \]
This is a quadratic function in terms of \( y \).

### Step 2: Find the critical points
To find the maximum value of \( f(y) \), we first take the derivative with respect to \( y \):
\[ f'(y) = 2y - x \]
Setting the derivative equal to zero to find critical points:
\[ 2y - x = 0 \implies y = \frac{x}{2} \]

### Step 3: Determine the nature of the critical point
Next, we need to check the value of \( y = \frac{x}{2} \) against the bounds \( y = 0 \) and \( y = 1 \):
- If \( \frac{x}{2} < 0 \), then \( y = 0 \) is the maximum.
- If \( \frac{x}{2} > 1 \), then \( y = 1 \) is the maximum.
- If \( 0 \leq \frac{x}{2} \leq 1 \), then \( y = \frac{x}{2} \) is the maximum.

### Step 4: Evaluate \( g(x) \) at the critical points
Now we evaluate \( g(x) \) at the relevant points:
1. At \( y = 0 \):
\[ g(x) = f(0) = 0^2 - x \cdot 0 = 0 \]
2. At \( y = 1 \):
\[ g(x) = f(1) = 1^2 - x \cdot 1 = 1 - x \]
3. At \( y = \frac{x}{2} \) (only valid if \( 0 \leq x \leq 2 \)):
\[ g(x) = f\left(\frac{x}{2}\right) = \left(\frac{x}{2}\right)^2 - x \cdot \frac{x}{2} = \frac{x^2}{4} - \frac{x^2}{2} = -\frac{x^2}{4} \]

### Step 5: Determine the maximum of these values
Now we need to find the maximum of \( g(x) \) for different ranges of \( x \):
- For \( x < 0 \), \( g(x) = 0 \).
- For \( 0 \leq x \leq 2 \), we compare:
- \( g(x) = 0 \) (at \( y = 0 \))
- \( g(x) = 1 - x \) (at \( y = 1 \))
- \( g(x) = -\frac{x^2}{4} \) (at \( y = \frac{x}{2} \))

### Step 6: Find the minimum value of \( g(x) \)
To find the minimum value of \( g(x) \) over all \( x \):
- For \( x = 0 \), \( g(0) = 0 \).
- For \( x = 1 \), \( g(1) = 0 \).
- For \( x = 2 \), \( g(2) = 0 \).

Thus, the minimum value of \( g(x) \) for real \( x \) is:
\[ \boxed{0} \]