a) Find the area bounded by the curve y^2 = 4x , the x axis and the line x = 4 . [2] b) The demand and supply function in a competitive market are given by the equation Q d = 240 3p and Q s = 5P 150…

a) Area bounded by the curve $y^2=4x$, the x-axis, and the line $x=4$

Step 1: Sketch the region The curve $y^2=4x$ is a right-opening parabola symmetric about the x-axis. The x-axis ($y=0$) and the vertical line $x=4$ bound the region.

Step 2: Find intersection points

• With the x-axis ($y=0$): $0^2=4x⟹x=0$. So, the parabola intersects the x-axis at $(0,0)$.
• With $x=4$: $y^2=16⟹y=\pm 4$. So, the vertical line intersects the parabola at $(4,4)$ and $(4,-4)$.

Step 3: Set up the integral The region is symmetric about the x-axis. We compute the area above the x-axis and double it. For $y\ge 0$, solve for $x$: $x=\frac{y^2}{4}$. The limits for $y$ are from $0$ to $4$ (since $y=4$ at $x=4$).

Area above x-axis: $$A_{\text{upper}}={\int }_0^{4}xdy={\int }_0^4\frac{y^2}{4}dy=\frac{1}{4}{\int }_0^4{y}^{2}dy$$

Step 4: Compute the integral $$A_{\text{upper}}=\frac{1}{4}{\left[\frac{y^3}{3}\right]}_0^4=\frac{1}{4}\left(\frac{64}{3}-0\right)=\frac{16}{3}$$

Step 5: Total area $$A_{\text{total}}=2\times A_{\text{upper}}=2\times \frac{16}{3}=\frac{32}{3}\approx 10.\overline{6}$$

Final answer: The area is $\frac{32}{3}$ square units.

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b) Solving the differential equation for price adjustment

Step 1: Write the given equations

• Demand: $Q_d=240-3p$
• Supply: $Q_s=5p-150$
• Differential equation: $\frac{dp}{dt}=0.05(Q_d-Q_s)$

Step 2: Compute $Q_d-Q_s$ $$Q_d-Q_s=(240-3p)-(5p-150)=390-8p$$ So, $$\frac{dp}{dt}=0.05(390-8p)=19.5-0.4p$$

Step 3: Rewrite as a separable ODE $$\frac{dp}{19.5-0.4p}=dt$$

Step 4: Integrate both sides Let $u=19.5-0.4p$, then $du=-0.4dp⟹dp=-\frac{du}{0.4}$. Substitute: $$\int \frac{dp}{19.5-0.4p}=\int dt⟹-\frac{1}{0.4}\ln |19.5-0.4p|=t+C$$ Simplify: $$-2.5\ln |19.5-0.4p|=t+C$$

Step 5: Solve for $p$ $$\ln |19.5-0.4p|=-\frac{t}{2.5}-\frac{C}{2.5}$$ Exponentiate: $$19.5-0.4p=e^{-\frac{t}{2.5}-\frac{C}{2.5}}=A{e}^{-\frac{t}{2.5}}$$ where $A=e^{-\frac{C}{2.5}}$.

Step 6: Apply initial condition $p(0)=50$ $$19.5-0.4(50)=A⟹19.5-20=A⟹A=-0.5$$ Thus: $$19.5-0.4p=-0.5e^{-\frac{t}{2.5}}$$ Solve for $p$: $$0.4p=19.5+0.5e^{-\frac{t}{2.5}}⟹p=\frac{19.5}{0.4}+\frac{0.5}{0.4}{e}^{-\frac{t}{2.5}}$$ $$p=48.75+1.25e^{-\frac{t}{2.5}}$$

Final answer: The price as a function of time is $$\boxed{p(t)=48.75+1.25e^{-\frac{t}{2.5}}}$$