If `x^(2) + 9y^(2) = 1`, then minimum and maximum value of `3x^(2) - 27y^(2) + 24xy` respectively

Options

A. `0, 5`
B. `-5, 5`
C. `-5, 10`
D. `0, 10`

Explanation

To find the minimum and maximum values of the expression \(3x^2 - 27y^2 + 24xy\) given the constraint \(x^2 + 9y^2 = 1\), we can use a trigonometric substitution.

### Step-by-Step Solution:

1. **Substitution**:
Let \(x = \cos \theta\). Then, substituting \(x\) into the constraint:
\[ x^2 + 9y^2 = 1 \implies \cos^2 \theta + 9y^2 = 1 \]
Rearranging gives:
\[ 9y^2 = 1 - \cos^2 \theta \implies 9y^2 = \sin^2 \theta \implies y^2 = \frac{\sin^2 \theta}{9} \implies y = \frac{1}{3} \sin \theta \]

2. **Substituting \(x\) and \(y\) into the expression**:
Now substitute \(x\) and \(y\) into the expression \(3x^2 - 27y^2 + 24xy\):
\[ 3x^2 = 3\cos^2 \theta \]
\[ -27y^2 = -27 \left(\frac{1}{9} \sin^2 \theta\right) = -3\sin^2 \theta \]
\[ 24xy = 24\left(\cos \theta \cdot \frac{1}{3} \sin \theta\right) = 8\sin \theta \cos \theta \]
Combining these gives:
\[ z = 3\cos^2 \theta - 3\sin^2 \theta + 8\sin \theta \cos \theta \]

3. **Using Trigonometric Identities**:
Using the identities \(\cos^2 \theta - \sin^2 \theta = \cos 2\theta\) and \(2\sin \theta \cos \theta = \sin 2\theta\):
\[ z = 3(\cos^2 \theta - \sin^2 \theta) + 4\sin 2\theta = 3\cos 2\theta + 4\sin 2\theta \]

4. **Finding Maximum and Minimum Values**:
The expression \(z = 3\cos 2\theta + 4\sin 2\theta\) can be rewritten in the form \(R\cos(2\theta - \phi)\), where:
\[ R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]
The angle \(\phi\) can be found using:
\[ \tan \phi = \frac{4}{3} \]
Thus, the maximum value of \(z\) is \(5\) and the minimum value is \(-5\).

5. **Final Values**:
Therefore, the minimum value of \(z\) is \(-5\) and the maximum value is \(5\).

### Conclusion:
The minimum value of \(3x^2 - 27y^2 + 24xy\) is \(-5\) and the maximum value is \(5\).