Question

Let the area of the region \(\{(x, y) : x - 2y + 4 \geq 0, x + 2y^2 \geq 0, x + 4y^2 \leq 8, y \geq 0\}\) be \(\frac{m}{n}\), where \( m \) and \( n \) are coprime numbers. Then \( m + n \) is equal to ______.

Answer 1

Correct Answer: 119

The region is defined by the inequalities:

\[ x - 2y + 4 \ge 0,\quad x + 2y^2 \ge 0,\quad x + 4y^2 \le 8,\quad y \ge 0 \]

The area \(A\) of this region is calculated as the sum of two integrals:

\[ A = \int_0^1 \left[(8 - 4y^2) - (-2y^2)\right] dy + \int_1^{3/2} \left[(8 - 4y^2) - (2y - 4)\right] dy \]

Integral 1 Evaluation

The first integral is:

\[ \int_0^1 \left[(8 - 4y^2) - (-2y^2)\right] dy = \int_0^1 (8 - 2y^2) \, dy \] \[ = \left[ 8y - \frac{2y^3}{3} \right]_0^1 = 8 - \frac{2}{3} = \frac{22}{3} \]

Integral 2 Evaluation

The second integral is:

\[ \int_1^{3/2} \left[(8 - 4y^2) - (2y - 4)\right] dy = \int_1^{3/2} (12 - 2y - 4y^2) \, dy \] \[ = \left[ 12y - y^2 - \frac{4y^3}{3} \right]_1^{3/2} \] \[ = \left(18 - \frac{9}{4} - \frac{27}{6}\right) - \left(12 - 1 - \frac{4}{3}\right) = \frac{19}{12} \]

Total Area Calculation

The total area \(A\) is the sum of the two integrals:

\[ A = \frac{22}{3} + \frac{19}{12} = \frac{107}{12} \]

Given \( A = \frac{m}{n} = \frac{107}{12} \), we find \( m + n \):

\[ m + n = 107 + 12 = \boxed{119} \]

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Final Answer: \( m + n = 119 \)