1. **State the problem:** We have a line $l$ with equation $2y = x$ and a curve $C$ with equation $y = 2x - \frac{1}{8}x^3$. The region $R$ is bounded by $l$, $C$, and a vertical line $l_1$ which passes through the positive $x$-intercept of $C$. We need to find the inequalities defining $R$. 2. **Find the positive $x$-intercept of curve $C$:** Set $y=0$ in $y = 2x - \frac{1}{8}x^3$: $$0 = 2x - \frac{1}{8}x^3$$ Factor out $x$: $$x\left(2 - \frac{1}{8}x^2\right) = 0$$ So $x=0$ or $2 - \frac{1}{8}x^2 = 0$. Solve for $x$: $$2 = \frac{1}{8}x^2 \implies x^2 = 16 \implies x = \pm 4$$ Positive intercept is at $x=4$. 3. **Equation of line $l_1$:** Since $l_1$ is vertical and passes through $x=4$, its equation is: $$x = 4$$ 4. **Determine inequalities for $R$:** - The region is bounded on the left by line $l$: $2y = x \implies y = \frac{x}{2}$. - The region is bounded on the right by line $l_1$: $x = 4$. - The region is bounded above or below by curve $C$: $y = 2x - \frac{1}{8}x^3$. 5. **Find which curve is above or below between $x=0$ and $x=4$:** Check at $x=2$: - $y_l = \frac{2}{2} = 1$ - $y_C = 2(2) - \frac{1}{8}(2)^3 = 4 - \frac{1}{8} \times 8 = 4 - 1 = 3$ Since $y_C > y_l$ at $x=2$, curve $C$ is above line $l$ in this interval. 6. **Inequalities defining $R$:** - $x$ is between $0$ and $4$: $0 \leq x \leq 4$ - $y$ is between line $l$ and curve $C$: $\frac{x}{2} \leq y \leq 2x - \frac{1}{8}x^3$ **Final answer:** $$\boxed{0 \leq x \leq 4, \quad \frac{x}{2} \leq y \leq 2x - \frac{1}{8}x^3}$$