Find the least and greatest value of `f(x,y) = x^(2) + y ^(2) - xy` where x and y are connected by the relation `x^(2)+4y^(2) = 4`.

Answer

To find the least and greatest values of the function \( f(x,y) = x^2 + y^2 - xy \) under the constraint \( x^2 + 4y^2 = 4 \), we can follow these steps:

### Step 1: Rewrite the constraint in standard form
The given constraint can be rewritten as:
\[ \frac{x^2}{4} + \frac{y^2}{1} = 1 \]
This represents an ellipse with semi-major axis \( a = 2 \) and semi-minor axis \( b = 1 \).

### Step 2: Use parametric equations for the ellipse
To express \( x \) and \( y \) in terms of a parameter \( \theta \), we use the parametric equations:
\[ x = 2 \cos \theta, \quad y = \sin \theta \]

### Step 3: Substitute the parametric equations into the function
Now substitute \( x \) and \( y \) into the function \( f(x,y) \):
\[ f(2 \cos \theta, \sin \theta) = (2 \cos \theta)^2 + (\sin \theta)^2 - (2 \cos \theta)(\sin \theta) \]
Calculating this gives:
\[ = 4 \cos^2 \theta + \sin^2 \theta - 2 \cos \theta \sin \theta \]
\[ = 4 \cos^2 \theta + \sin^2 \theta - \sin(2\theta) \]

### Step 4: Simplify the expression
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ = 4 \cos^2 \theta + (1 - \cos^2 \theta) - \sin(2\theta) \]
\[ = 3 \cos^2 \theta + 1 - \sin(2\theta) \]

### Step 5: Express \( \sin(2\theta) \) in terms of \( \cos(2\theta) \)
Recall that \( \sin(2\theta) = 2 \sin \theta \cos \theta \). We can also express \( \cos^2 \theta \) in terms of \( \cos(2\theta) \):
\[ \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} \]
Substituting this back, we get:
\[ f(\theta) = 3 \left(\frac{1 + \cos(2\theta)}{2}\right) + 1 - \sin(2\theta) \]
\[ = \frac{3}{2} + \frac{3}{2} \cos(2\theta) + 1 - \sin(2\theta) \]
\[ = \frac{5}{2} + \frac{3}{2} \cos(2\theta) - \sin(2\theta) \]

### Step 6: Find the maximum and minimum values
This expression can be treated as \( A \cos(2\theta) + B \sin(2\theta) \) where \( A = \frac{3}{2} \) and \( B = -1 \). The maximum and minimum values of \( A \cos(2\theta) + B \sin(2\theta) \) can be found using:
\[ \sqrt{A^2 + B^2} = \sqrt{\left(\frac{3}{2}\right)^2 + (-1)^2} = \sqrt{\frac{9}{4} + 1} = \sqrt{\frac{13}{4}} = \frac{\sqrt{13}}{2} \]

Thus, the maximum and minimum values of the function are:
\[ \text{Maximum} = \frac{5}{2} + \frac{\sqrt{13}}{2} \]
\[ \text{Minimum} = \frac{5}{2} - \frac{\sqrt{13}}{2} \]

### Final Answer
The least value of \( f(x,y) \) is \( \frac{5 - \sqrt{13}}{2} \) and the greatest value is \( \frac{5 + \sqrt{13}}{2} \).

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