© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Understand the Given Inequalities
We have the following set of inequalities describing a region in the xy -plane:
1. x - 2y + 4 \geq 0 \quad \Longrightarrow \quad x \geq 2y - 4
2. x + 2y^2 \geq 0 \quad \Longrightarrow \quad x \geq -2y^2
3. x + 4y^2 \leq 8 \quad \Longrightarrow \quad x \leq 8 - 4y^2
4. y \geq 0
For any fixed y \ge 0, the value of x must lie between the larger of the two lower bounds ( 2y - 4 and -2y^2 ) and the upper bound 8 - 4y^2 .
Step 2: Find the Effective Lower Bound for x
Compare 2y - 4 and -2y^2 to determine which is larger at different y values. We set them equal to find the point of intersection:
2y - 4 = -2y^2
\quad\Longrightarrow\quad
2y^2 + 2y - 4 = 0
\quad\Longrightarrow\quad
y^2 + y - 2 = 0.
Solving this quadratic:
y = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm 3}{2}.
The roots are y = 1 and y = -2. Since y \ge 0, the relevant intersection point is y = 1.
• For 0 \le y < 1, we have -2y^2 > 2y - 4, so x \ge -2y^2 is the stricter lower bound.
• For y \ge 1, we have 2y - 4 \ge -2y^2, so x \ge 2y - 4 is the stricter lower bound.
Step 3: Determine the Valid Range for y
The upper bound for x is 8 - 4y^2. We need this to be at least as large as the stricter lower bound in each region:
Region 1: 0 \le y \le 1 .
The x -range is -2y^2 \le x \le 8 - 4y^2.
Region 2: 1 \le y \leq ? .
The x -range is 2y - 4 \le x \le 8 - 4y^2.
We impose 2y - 4 \le 8 - 4y^2 to find the maximum y where this holds:
2y - 4 \le 8 - 4y^2
\quad\Longrightarrow\quad
4y^2 + 2y - 12 \le 0
\quad\Longrightarrow\quad
2y^2 + y - 6 \le 0.
Solving 2y^2 + y - 6 = 0 :
y = \frac{-1 \pm \sqrt{1 + 48}}{4} = \frac{-1 \pm 7}{4}.
The positive root is y = \frac{6}{4} = 1.5.
Hence for 1 \le y \le 1.5, the x -range is 2y - 4 \le x \le 8 - 4y^2.
Step 4: Divide and Calculate the Area in Two Parts
We will find the area in two separate intervals for y and then add them:
Part A: 0 \le y \le 1
The width in x is:
(8 - 4y^2) - (-2y^2) = 8 - 4y^2 + 2y^2 = 8 - 2y^2.
Therefore, \text{Area}_A is:
\int_{0}^{1} \bigl(8 - 2y^2\bigr)\, dy
= \left[\,8y - \frac{2y^3}{3}\right]_{0}^{1}
= \left(8 - \frac{2}{3}\right) - 0
= 8 - \frac{2}{3}
= \frac{24 - 2}{3}
= \frac{22}{3}.
Part B: 1 \le y \le 1.5
The width in x is:
(8 - 4y^2) - (2y - 4)
= 8 - 4y^2 - 2y + 4
= 12 - 4y^2 - 2y.
Hence, \text{Area}_B is:
\int_{1}^{1.5} \bigl(12 - 4y^2 - 2y\bigr)\, dy.
Define
F(y) = \int \bigl(12 - 4y^2 - 2y\bigr)\, dy
= 12y - \frac{4y^3}{3} - y^2.
We need F(1.5) - F(1) :
Calculate F(1) :
F(1) = 12(1) - \frac{4 \cdot 1^3}{3} - 1^2
= 12 - \frac{4}{3} - 1
= 11 - \frac{4}{3}
= \frac{33 - 4}{3}
= \frac{29}{3}.
Calculate F(1.5) :
Since (1.5)^3 = 3.375,
\frac{4 \times 3.375}{3} = \frac{13.5}{3} = 4.5,
12 \times 1.5 = 18, \quad 18 - 4.5 = 13.5, \quad (1.5)^2 = 2.25, \quad 13.5 - 2.25 = 11.25.
In fractional form, 11.25 = \frac{45}{4}. Thus
F(1.5) = \frac{45}{4}.
Therefore,
\bigl[F(1.5) - F(1)\bigr]
= \frac{45}{4} - \frac{29}{3}
= \frac{135}{12} - \frac{116}{12}
= \frac{19}{12}.
Hence, \text{Area}_B = \frac{19}{12}.
Step 5: Total Area
The total area is the sum of these two portions:
\text{Area} = \text{Area}_A + \text{Area}_B
= \frac{22}{3} + \frac{19}{12}
= \frac{88}{12} + \frac{19}{12}
= \frac{107}{12}.
Step 6: Final Answer
According to the problem statement, the area is \frac{m}{n} in lowest terms. We have
\frac{m}{n} = \frac{107}{12}.
Since 107 and 12 are coprime, m = 107 and n = 12. Thus,
m + n = 107 + 12 = 119.
Therefore, the value of m + n is
119.
