Let $$f : [0, 1] \to [0, 1]$$ be the function defined by $$f(x) = \frac{x^3}{3} - x^2 + \frac{5}{9}x + \frac{17}{36}$$. Consider the square region $$S = [0, 1] \times [0, 1]$$. Let $$G = \{(x, y) \in S : y > f(x)\}$$ be called the green region and $$R = \{(x, y) \in S : y < f(x)\}$$ be called the red region. Let $$L_h = \{(x, h) \in S : x \in [0, 1]\}$$ be the horizontal line drawn at a height $$h \in [0, 1]$$. Then which of the following statements is(are) true?

Let $$f:[0,1]\rightarrow[0,1]$$ be given by $$f(x)=\dfrac{x^{3}}{3}-x^{2}+\dfrac{5}{9}x+\dfrac{17}{36}$$ and let the unit square be $$S=[0,1]\times[0,1]$$.

Throughout the solution we denote by $$G_{\,\text{above}}(h)=\text{area of }\{(x,y)\in S:y\gt h\text{ and }y\gt f(x)\}$$ $$G_{\,\text{below}}(h)=\text{area of }\{(x,y)\in S:y\lt h\text{ and }y\gt f(x)\}$$ $$R_{\,\text{above}}(h)=\text{area of }\{(x,y)\in S:y\gt h\text{ and }y\lt f(x)\}$$ $$R_{\,\text{below}}(h)=\text{area of }\{(x,y)\in S:y\lt h\text{ and }y\lt f(x)\}$$

All four quantities depend continuously on $$h\in[0,1]$$, because the integrands that define them are continuous in $$h$$.

Step 1: Total red area and total green area
For every fixed $$x\in[0,1]$$ the red part occupies the vertical segment $$0\le y\le f(x)$$ and the green part the segment $$f(x)\le y\le 1$$. Hence
$$\text{(total red area)}=\int_{0}^{1}f(x)\,dx,\qquad \text{(total green area)}=1-\int_{0}^{1}f(x)\,dx.$$ Compute the integral:

$$\int_{0}^{1}f(x)\,dx =\int_{0}^{1}\!\left(\frac{x^{3}}{3}-x^{2}+\frac{5}{9}x+\frac{17}{36}\right)\!dx =\frac{1}{12}-\frac{1}{3}+\frac{5}{18}+\frac{17}{36} =\frac{18}{36}=\frac12.$$

Therefore
$$\boxed{\text{Total red area}= \dfrac12},\qquad \boxed{\text{Total green area}= \dfrac12}.$$

Step 2: Minimum and maximum values of $$f(x)$$
The derivative is $$f'(x)=x^{2}-2x+\dfrac59=(x-\tfrac13)(x-\tfrac53).$$ The only critical point in $$[0,1]$$ is $$x=\tfrac13$$, and because $$f''(x)=2x-2$$, $$f''(\tfrac13)=-\dfrac43\lt0$$, that point is a maximum. Consequently the minimum occurs at one of the end-points:

$$f(0)=\frac{17}{36}=0.4722\ldots,\qquad f(1)=\frac13-1+\frac59+\frac{17}{36}=0.3611\ldots.$$ Hence $$\boxed{0.361\lt f(x)\lt 0.559\;\text{ for every }x\in[0,1]}.$$ In particular $$f(x)\gt\dfrac14$$ and $$f(x)\lt\dfrac23$$ for every $$x\in[0,1]$$. These facts will be used repeatedly.

Step 3: Red-region equality $$R_{\,\text{above}}(h)=R_{\,\text{below}}(h)$$ (Option B)
For each $$h$$ $$R_{\,\text{above}}(h)=\int_{0}^{1}\max\!\bigl(f(x)-h,0\bigr)\,dx,$$ $$R_{\,\text{below}}(h)=\int_{0}^{1}\min\!\bigl(f(x),h\bigr)\,dx.$$ If $$f(x)\ge h$$ the first integrand is $$f(x)-h$$ and the second $$h$$; if $$f(x)\le h$$ the first is $$0$$ and the second $$f(x)$$. Hence for every $$x$$ $$\max(f(x)-h,0)+\min(f(x),h)=f(x)$$ which gives $$R_{\,\text{above}}(h)+R_{\,\text{below}}(h)=\int_{0}^{1}f(x)\,dx=\frac12.$$ We therefore need $$R_{\,\text{above}}(h)=\dfrac14.$$ Define $$\phi(h)=R_{\,\text{above}}(h).$$ Because $$\phi(0)=\tfrac12$$ and $$\phi(1)=0$$, by the Intermediate Value Theorem there exists some $$h$$ with $$\phi(h)=\tfrac14$$. The choice $$h=\dfrac14$$ itself works, because $$f(x)\gt\dfrac14$$ for every $$x$$, so $$R_{\,\text{above}}\!\bigl(\tfrac14\bigr)=\int_{0}^{1}\!(f(x)-\tfrac14)\,dx =\frac12-\frac14=\frac14.$$ Thus a suitable $$h$$ lies in $$\bigl[\tfrac14,\tfrac23\bigr]$$, proving Option B is true.

Step 4: Green-above equals Red-below $$G_{\,\text{above}}(h)=R_{\,\text{below}}(h)$$ (Option C)
Write $$\Phi(h)=G_{\,\text{above}}(h)-R_{\,\text{below}}(h).$$ Using the facts $$G_{\,\text{above}}(h)+R_{\,\text{above}}(h)=1-h,\qquad G_{\,\text{below}}(h)+R_{\,\text{below}}(h)=h,$$ we still evaluate directly at the two end-points of the interval in the option:

• For $$h=\dfrac14$$, since $$f(x)\gt\dfrac14$$ for all $$x$$, $$G_{\,\text{above}}\!\bigl(\tfrac14\bigr)=\int_{0}^{1}(1-f(x))\,dx=\frac12,$$ $$R_{\,\text{below}}\!\bigl(\tfrac14\bigr)=\int_{0}^{1}\tfrac14\,dx=\tfrac14,$$ so $$\Phi\!\bigl(\tfrac14\bigr)=\tfrac12-\tfrac14=\tfrac14\gt0.$$

• For $$h=\dfrac23$$, because $$f(x)\lt\dfrac23$$ for every $$x$$, $$G_{\,\text{above}}\!\bigl(\tfrac23\bigr)=\int_{0}^{1}(1-\tfrac23)\,dx=\tfrac13,$$ $$R_{\,\text{below}}\!\bigl(\tfrac23\bigr)=\int_{0}^{1}f(x)\,dx=\tfrac12,$$ so $$\Phi\!\bigl(\tfrac23\bigr)=\tfrac13-\tfrac12=-\tfrac16\lt0.$$

Because $$\Phi(h)$$ is continuous and changes sign between $$h=\tfrac14$$ and $$h=\tfrac23$$, there exists some $$h\in\bigl(\tfrac14,\tfrac23\bigr)$$ with $$\Phi(h)=0,$$ establishing Option C as true.

Step 5: Red-above equals Green-below $$R_{\,\text{above}}(h)=G_{\,\text{below}}(h)$$ (Option D)
Put $$\Psi(h)=R_{\,\text{above}}(h)-G_{\,\text{below}}(h).$$ A direct calculation shows $$\Phi(h)+\Psi(h)=(1-h)-h=1-2h.$$ We already know $$\Phi(h)$$ is continuous; hence so is $$\Psi(h)$$. Again evaluate at the two ends:

• $$h=\dfrac14:\qquad R_{\,\text{above}}=\tfrac14,\; G_{\,\text{below}}=0\;\Longrightarrow\; \Psi\bigl(\tfrac14\bigr)=\tfrac14\gt0.$$

• $$h=\dfrac23:\qquad R_{\,\text{above}}=0,\; G_{\,\text{below}}=\tfrac16\;\Longrightarrow\; \Psi\bigl(\tfrac23\bigr)=-\tfrac16\lt0.$$

Since $$\Psi(h)$$ changes sign on $$\bigl[\tfrac14,\tfrac23\bigr]$$, an $$h$$ in this interval satisfies $$\Psi(h)=0$$, proving Option D is true.

Step 6: Green-above equals Green-below (Option A)
For the green region we want $$G_{\,\text{above}}(h)=G_{\,\text{below}}(h).$$ Because the total green area is $$\tfrac12$$, the equality is equivalent to $$G_{\,\text{above}}(h)=\tfrac14.$$ Define $$\gamma(h)=G_{\,\text{above}}(h).$$ We have already obtained $$\gamma(0)=\tfrac12,\qquad \gamma\!\bigl(\tfrac23\bigr)=\tfrac13.$$ Thus on $$[0,\tfrac23]$$ the value of $$\gamma(h)$$ stays at least $$\tfrac13\gt\tfrac14,$$ so it can equal $$\tfrac14$$ only for some $$h\gt\tfrac23$$. Consequently no $$h$$ in $$\bigl[\tfrac14,\tfrac23\bigr]$$ satisfies the required condition, and Option A is false.

Conclusion
The correct statements are:
Option B, Option C and Option D.

Hence the answer is:
Option B (red above = red below), Option C (green above = red below) and Option D (red above = green below).