1. **State the problem:** We have the curve $$y = x^3 - 4x^{5/2} - kx^{1/2} + 28x - 44$$ for $$x \geq 0$$, where $$k$$ is a positive constant. (a) Find $$\frac{dy}{dx}$$ in simplest form. (b) Given that the point $$T$$ is a minimum stationary point with $$x=9$$, show that $$k=6$$. (c) Find the area of the shaded region $$R$$ bounded by the curve, the y-axis, and the horizontal line through $$T$$. --- 2. **Find $$\frac{dy}{dx}$$:** Use the power rule $$\frac{d}{dx} x^n = nx^{n-1}$$. $$y = x^3 - 4x^{5/2} - kx^{1/2} + 28x - 44$$ Differentiate term-by-term: $$\frac{dy}{dx} = 3x^2 - 4 \times \frac{5}{2} x^{\frac{5}{2} - 1} - k \times \frac{1}{2} x^{\frac{1}{2} - 1} + 28 - 0$$ Simplify exponents: $$\frac{dy}{dx} = 3x^2 - 10 x^{3/2} - \frac{k}{2} x^{-1/2} + 28$$ --- 3. **Use the stationary point condition at $$x=9$$:** At stationary points, $$\frac{dy}{dx} = 0$$. Substitute $$x=9$$: $$0 = 3(9)^2 - 10 (9)^{3/2} - \frac{k}{2} (9)^{-1/2} + 28$$ Calculate powers: $$9^2 = 81$$ $$9^{3/2} = (\sqrt{9})^3 = 3^3 = 27$$ $$9^{-1/2} = \frac{1}{\sqrt{9}} = \frac{1}{3}$$ Substitute: $$0 = 3 \times 81 - 10 \times 27 - \frac{k}{2} \times \frac{1}{3} + 28$$ Simplify: $$0 = 243 - 270 - \frac{k}{6} + 28$$ Combine constants: $$243 - 270 + 28 = 1$$ So: $$0 = 1 - \frac{k}{6}$$ Rearranged: $$\frac{k}{6} = 1 \implies k = 6$$ --- 4. **Find the y-coordinate of point $$T$$:** Substitute $$x=9$$ and $$k=6$$ into $$y$$: $$y = 9^3 - 4 \times 9^{5/2} - 6 \times 9^{1/2} + 28 \times 9 - 44$$ Calculate powers: $$9^3 = 729$$ $$9^{5/2} = (\sqrt{9})^5 = 3^5 = 243$$ $$9^{1/2} = 3$$ Substitute: $$y = 729 - 4 \times 243 - 6 \times 3 + 252 - 44$$ Calculate terms: $$-4 \times 243 = -972$$ $$-6 \times 3 = -18$$ Sum all: $$729 - 972 - 18 + 252 - 44 = (729 - 972) + ( -18 + 252 - 44) = -243 + 190 = -53$$ So $$T = (9, -53)$$. --- 5. **Find the y-intercept of the horizontal line through $$T$$:** The line through $$T$$ parallel to the x-axis has equation $$y = -53$$. At the y-axis, $$x=0$$, so point $$N = (0, -53)$$. --- 6. **Find the area of region $$R$$ bounded by curve $$C$$, y-axis, and line segment $$NT$$:** Area $$= \int_0^9 y \, dx - \text{area under line } y=-53$$ Since the line is below the curve, the area between curve and line is: $$\text{Area} = \int_0^9 \bigl[y - (-53)\bigr] dx = \int_0^9 (y + 53) dx$$ Substitute $$y$$: $$\int_0^9 \left(x^3 - 4x^{5/2} - 6x^{1/2} + 28x - 44 + 53\right) dx = \int_0^9 \left(x^3 - 4x^{5/2} - 6x^{1/2} + 28x + 9\right) dx$$ Integrate term-by-term: $$\int_0^9 x^3 dx = \left[ \frac{x^4}{4} \right]_0^9 = \frac{9^4}{4} = \frac{6561}{4} = 1640.25$$ $$\int_0^9 -4x^{5/2} dx = -4 \left[ \frac{x^{7/2}}{7/2} \right]_0^9 = -4 \times \frac{2}{7} x^{7/2} \Big|_0^9 = -\frac{8}{7} 9^{7/2}$$ Calculate $$9^{7/2} = (\sqrt{9})^7 = 3^7 = 2187$$ So: $$-\frac{8}{7} \times 2187 = -\frac{17496}{7} \approx -2499.43$$ $$\int_0^9 -6x^{1/2} dx = -6 \left[ \frac{2}{3} x^{3/2} \right]_0^9 = -6 \times \frac{2}{3} 9^{3/2} = -4 \times 27 = -108$$ $$\int_0^9 28x dx = 28 \left[ \frac{x^2}{2} \right]_0^9 = 28 \times \frac{81}{2} = 28 \times 40.5 = 1134$$ $$\int_0^9 9 dx = 9x \Big|_0^9 = 81$$ Sum all integrals: $$1640.25 - 2499.43 - 108 + 1134 + 81 = (1640.25 + 1134 + 81) - (2499.43 + 108) = 2855.25 - 2607.43 = 247.82$$ --- **Final answers:** (a) $$\frac{dy}{dx} = 3x^2 - 10x^{3/2} - \frac{k}{2} x^{-1/2} + 28$$ (b) $$k = 6$$ (c) Area of region $$R \approx 248$$ (to 3 significant figures)