Area Division of Green and Red Regions by Horizontal Line

To determine which of the given statements are true, let us first analyze the properties of the function $f(x) = \frac{x^3}{3} - x^2 + \frac{5}{9}x + \frac{17}{36}$ on the interval $[0,1]$.

1. Extrema and Range of $f(x)$

Taking the derivative of $f(x)$: $f'(x) = x^2 - 2x + \frac{5}{9}$

Setting $f'(x) = 0$: $x^2 - 2x + \frac{5}{9} = 0 \implies (x - 1)^2 = 1 - \frac{5}{9} = \frac{4}{9} \implies x = 1 \pm \frac{2}{3}$

The critical point inside $[0,1]$ is $x = \frac{1}{3}$. Evaluating $f(x)$ at the critical point and boundary points:

• At $x = 0$: $f(0) = \frac{17}{36}$

• At $x = \frac{1}{3}$: $f\left(\frac{1}{3}\right) = \frac{1}{81} - \frac{1}{9} + \frac{5}{27} + \frac{17}{36} = \frac{181}{324}$

• At $x = 1$: $f(1) = \frac{1}{3} - 1 + \frac{5}{9} + \frac{17}{36} = \frac{13}{36}$

Note that: $\frac{1}{4} = \frac{9}{36} < \frac{13}{36} \le f(x) \le \frac{181}{324} < \frac{216}{324} = \frac{2}{3}$

Thus, for all $x \in [0,1]$, we have $f(x) \in \left[\frac{13}{36}, \frac{181}{324}\right] \subset \left(\frac{1}{4}, \frac{2}{3}\right)$.

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2. Area Calculation of the Regions

The total area of the square region $S = [0,1] \times [0,1]$ is $\text{Area}(S) = 1$.

The area of the red region $R = \{(x,y) \in S : y < f(x)\}$ is given by: $\text{Area}(R) = \int_0^1 f(x) \, dx = \left[ \frac{x^4}{12} - \frac{x^3}{3} + \frac{5x^2}{18} + \frac{17x}{36} \right]_0^1$ $\text{Area}(R) = \frac{1}{12} - \frac{1}{3} + \frac{5}{18} + \frac{17}{36} = \frac{3 - 12 + 10 + 17}{36} = \frac{18}{36} = \frac{1}{2}$

Since the total area of $S$ is $1$, the area of the green region $G = \{(x,y) \in S : y > f(x)\}$ is: $\text{Area}(G) = 1 - \text{Area}(R) = 1 - \frac{1}{2} = \frac{1}{2}$

For a horizontal line $L_h$ at height $h \in [0,1]$:

• $\text{Area of } S \text{ above } L_h = 1 - h$
• $\text{Area of } S \text{ below } L_h = h$

Thus, we have the following relations:

1. $G_{\text{above}}(h) + R_{\text{above}}(h) = 1 - h$
2. $G_{\text{below}}(h) + R_{\text{below}}(h) = h$
3. $G_{\text{above}}(h) + G_{\text{below}}(h) = \frac{1}{2}$
4. $R_{\text{above}}(h) + R_{\text{below}}(h) = \frac{1}{2}$

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3. Analysis of Options

Option A:

We check if there exists $h \in \left[\frac{1}{4}, \frac{2}{3}\right]$ such that $G_{\text{above}}(h) = G_{\text{below}}(h)$.

Since $f(x) < \frac{2}{3}$ for all $x \in [0,1]$, at $h = \frac{2}{3}$, the line $L_{2/3}$ lies completely above the graph of $f(x)$.

• $G_{\text{above}}\left(\frac{2}{3}\right) = 1 - \frac{2}{3} = \frac{1}{3}$
• $G_{\text{below}}\left(\frac{2}{3}\right) = \text{Area}(G) - G_{\text{above}}\left(\frac{2}{3}\right) = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$

For $h \ge \max f(x) = \frac{181}{324}$, we have $G_{\text{above}}(h) = 1 - h$. Setting $G_{\text{above}}(h) = G_{\text{below}}(h) \implies 1 - h = \frac{1}{4} \implies h = \frac{3}{4}$.

Since $\frac{3}{4} \notin \left[\frac{1}{4}, \frac{2}{3}\right]$, no such $h$ exists in the interval $\left[\frac{1}{4}, \frac{2}{3}\right]$. Hence, Option A is False.

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Option B:

We check if there exists $h \in \left[\frac{1}{4}, \frac{2}{3}\right]$ such that $R_{\text{above}}(h) = R_{\text{below}}(h)$.

Since $f(x) \ge \frac{13}{36} > \frac{1}{4}$ for all $x \in [0,1]$, the line $L_{1/4}$ lies entirely below the graph of $f(x)$. Therefore, the region below $L_{1/4}$ lies entirely within the red region $R$: $R_{\text{below}}\left(\frac{1}{4}\right) = \frac{1}{4} \times 1 = \frac{1}{4}$

Since $\text{Area}(R) = \frac{1}{2}$: $R_{\text{above}}\left(\frac{1}{4}\right) = \text{Area}(R) - R_{\text{below}}\left(\frac{1}{4}\right) = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$

Thus, $R_{\text{above}}\left(\frac{1}{4}\right) = R_{\text{below}}\left(\frac{1}{4}\right)$ at $h = \frac{1}{4} \in \left[\frac{1}{4}, \frac{2}{3}\right]$. Hence, Option B is True.

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Option C:

We check if there exists $h \in \left[\frac{1}{4}, \frac{2}{3}\right]$ such that $G_{\text{above}}(h) = R_{\text{below}}(h)$.

Using the relations: $G_{\text{above}}(h) + R_{\text{above}}(h) = 1 - h$ $R_{\text{below}}(h) + R_{\text{above}}(h) = \frac{1}{2}$

Subtracting these equations yields: $G_{\text{above}}(h) - R_{\text{below}}(h) = \frac{1}{2} - h$

Setting $G_{\text{above}}(h) = R_{\text{below}}(h) \implies \frac{1}{2} - h = 0 \implies h = \frac{1}{2}$. Since $h = \frac{1}{2} \in \left[\frac{1}{4}, \frac{2}{3}\right]$, such an $h$ exists. Hence, Option C is True.

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Option D:

We check if there exists $h \in \left[\frac{1}{4}, \frac{2}{3}\right]$ such that $R_{\text{above}}(h) = G_{\text{below}}(h)$.

Using the relations: $G_{\text{below}}(h) + R_{\text{below}}(h) = h$ $R_{\text{above}}(h) + R_{\text{below}}(h) = \frac{1}{2}$

Subtracting these equations yields: $R_{\text{above}}(h) - G_{\text{below}}(h) = \frac{1}{2} - h$

Setting $R_{\text{above}}(h) = G_{\text{below}}(h) \implies \frac{1}{2} - h = 0 \implies h = \frac{1}{2}$. Since $h = \frac{1}{2} \in \left[\frac{1}{4}, \frac{2}{3}\right]$, such an $h$ exists. Hence, Option D is True.

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Conclusion

The correct options are (B), (C), and (D).