Given that \( 1 \leqslant x \leqslant 3 \) and \( 3 \leqslant y \leqslant 6 \), find (a) the largest possible value of \( x^{2}-y \), (b) the smallest possible value of \( \frac{x}{y} \).

To solve the problems, we will analyze each part separately. ### Part (a): Find the largest possible value of \( x^{2} - y \) **Step 1**: Determine the range of \( x^{2} \). Given \( 1 \leq x \leq 3 \): - The minimum value of \( x^{2} \) occurs when \( x = 1 \): \[ x^{2} = 1^{2} = 1 \] - The maximum value of \( x^{2} \) occurs when \( x = 3 \): \[ x^{2} = 3^{2} = 9 \] So, the range of \( x^{2} \) is \( [1, 9] \). **Step 2**: Determine the range of \( y \). Given \( 3 \leq y \leq 6 \). **Step 3**: Find \( x^{2} - y \). Since we want to maximize \( x^{2} - y \), we first maximize \( x^{2} \) and minimize \( y \): - Maximum of \( x^{2} = 9 \) - Minimum of \( y = 3 \) Now, compute: \[ x^{2} - y = 9 - 3 = 6 \] **Answer for (a)**: The largest possible value of \( x^{2} - y \) is \( 6 \). ### Part (b): Find the smallest possible value of \( \frac{x}{y} \) **Step 1**: Determine the range of \( \frac{x}{y} \). We analyze \( \frac{x}{y} \) by minimizing \( x \) and maximizing \( y \). - Minimum of \( x = 1 \) - Maximum of \( y = 6 \) Compute: \[ \frac{x}{y} = \frac{1}{6} \] **Step 2**: Check for other combinations. Next, we also check the opposite scenario to ensure we find the minimum: - Maximum of \( x = 3 \) - Minimum of \( y = 3 \) Compute: \[ \frac{x}{y} = \frac{3}{3} = 1 \] **Conclusion**: The smallest value out of the calculated values \( \frac{1}{6} \) and \( 1 \) is \( \frac{1}{6} \). **Answer for (b)**: The smallest possible value of \( \frac{x}{y} \) is \( \frac{1}{6} \).