1. **Problem Statement (QUESTION ONE)**: A petroleum company operates two refineries. We want to formulate and solve its operating cost minimization problem subject to meeting oil production demands. 2. **Part (a): Objective Function** - Let $x$ = days refinery 1 operates - Let $y$ = days refinery 2 operates - Costs: refinery 1: 20000 per day, refinery 2: 25000 per day - Objective: minimize total cost $$ \text{Minimize } C = 20000x + 25000y $$ 3. **Part (b): Constraints** - Production constraints must meet or exceed demand: - High-grade oil: $400x + 300y \geq 25000$ - Medium-grade oil: $300x + 400y \geq 27000$ - Low-grade oil: $200x + 500y \geq 30000$ - Non-negativity: $x \geq 0, y \geq 0$ 4. **Part (c): Solve for $x$ and $y$ minimizing cost** We minimize $$C = 20000x + 25000y$$ subject to $$400x + 300y \geq 25000$$ $$300x + 400y \geq 27000$$ $$200x + 500y \geq 30000$$ $$x, y \geq 0$$ 5. **Convert inequalities to equalities to find candidate points:** - From pairs of constraints, we solve linear systems: (a) Solve $$400x + 300y = 25000$$ $$300x + 400y = 27000$$ Multiply first by 4 and second by 3 to equalize $y$'s coefficients: $$1600x + 1200y = 100000$$ $$900x + 1200y = 81000$$ Subtract second from first: $$700x = 19000 \implies x = \frac{19000}{700} = 27.14$$ Substitute $x$: $$400(27.14) + 300y = 25000 \implies 10856 + 300y = 25000 \implies 300y = 14144 \implies y = 47.15$$ Check third constraint: $$200(27.14) + 500(47.15) = 5428 + 23575 = 29003 < 30000$$ not satisfying third. 6. **Try constraints 1 and 3:** $$400x + 300y = 25000$$ $$200x + 500y = 30000$$ Multiply second by 2: $$400x + 1000y = 60000$$ Subtract first: $$400x + 1000y - 400x - 300y = 60000 - 25000 \implies 700y = 35000 \implies y = 50$$ Substitute in first: $$400x + 300(50) = 25000 \implies 400x + 15000 = 25000 \implies 400x = 10000 \implies x = 25$$ Check second constraint: $$300(25) + 400(50) = 7500 + 20000 = 27500 \geq 27000$$ Ok and both $x,y \geq 0$ 7. **Try constraints 2 and 3:** $$300x + 400y = 27000$$ $$200x + 500y = 30000$$ Multiply first by 2: $$600x + 800y = 54000$$ Multiply second by 3: $$600x + 1500y = 90000$$ Subtract first: $$600x + 1500y - (600x + 800y) = 90000 - 54000 \implies 700y = 36000 \implies y = 51.43$$ Substitute in first: $$300x + 400(51.43) = 27000 \implies 300x + 20572 = 27000 \implies 300x = 6430 \implies x = 21.43$$ Check first constraint: $$400(21.43) + 300(51.43) = 8572 + 15429 = 23991 < 25000$$ not satisfied. 8. **Feasible solution** from (6): $x=25$, $y=50$ Calculate cost: $$C = 20000(25) + 25000(50) = 500000 + 1250000 = 1750000$$ Check all constraints satisfied. 9. **Final answer for QUESTION ONE:** - Objective function: $C = 20000x + 25000y$ - Constraints: - $400x + 300y \geq 25000$ - $300x + 400y \geq 27000$ - $200x + 500y \geq 30000$ - $x,y \geq 0$ - Minimum cost solution: $x=25$, $y=50$, cost = 1750000 --- 10. **Problem Statement (QUESTION TWO)**: Find Taylor polynomials and expansions for given functions. 11. **Part (a): Taylor polynomials for $f(x) = \ln(x)$ at $x=1$** - Calculate derivatives: - $f(x)=\ln x$ - $f(1) = 0$ - $f'(x) = 1/x$, $f'(1) = 1$ - $f''(x) = -1/x^2$, $f''(1) = -1$ - $f'''(x) = 2/x^3$, $f'''(1) = 2$ 12. **Taylor polynomials:** - $p_0(x) = f(1) = 0$ - $p_1(x) = 0 + 1(x-1) = x - 1$ - $p_2(x) = p_1(x) + \frac{-1}{2!} (x-1)^2 = x - 1 - \frac{(x-1)^2}{2}$ - $p_3(x) = p_2 + \frac{2}{3!} (x-1)^3 = x -1 - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3}$ 13. **Graphing utility comparison:** compare $f(x)$ and $p_0, p_1, p_2, p_3$ around 1. 14. **Part (b): Expand $\sin(\pi/6 + h)$ to $h^4$ term** - Use Taylor series about $h=0$ for $\sin(a+h)$: $$\sin(a+h) = \sin a + h \cos a - \frac{h^2}{2} \sin a - \frac{h^3}{6} \cos a + \frac{h^4}{24} \sin a + \cdots$$ - Substitute $a=\pi/6$: $$\sin(\pi/6) = \frac{1}{2}, \cos(\pi/6) = \frac{\sqrt{3}}{2}$$ - Thus $$\sin(\pi/6 + h) = \frac{1}{2} + h \frac{\sqrt{3}}{2} - \frac{h^2}{2} \frac{1}{2} - \frac{h^3}{6} \frac{\sqrt{3}}{2} + \frac{h^4}{24} \frac{1}{2}$$ - Simplify: $$= \frac{1}{2} + \frac{\sqrt{3}}{2} h - \frac{h^2}{4} - \frac{\sqrt{3}}{12} h^3 + \frac{h^4}{48}$$ 15. **Part (c): Power series for $\cos^2 2x$ to $x^6$ term**: - Use identity: $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$ - So $$\cos^2 2x = \frac{1 + \cos 4x}{2}$$ Expand $\cos 4x$ using Maclaurin series: $$\cos 4x = 1 - \frac{(4x)^2}{2!} + \frac{(4x)^4}{4!} - \frac{(4x)^6}{6!} + \cdots = 1 - 8x^2 + \frac{(256)x^4}{24} - \frac{4096 x^6}{720}$$ Simplify coefficients: - $\frac{256}{24} = 10.6667$ - $\frac{4096}{720} = 5.6889$ So $$\cos^2 2x = \frac{1}{2} + \frac{1}{2} \left(1 - 8x^2 + 10.6667 x^4 - 5.6889 x^6 \right) = 1 - 4x^2 + 5.3333 x^4 - 2.8444 x^6$$ --- **Final answers:** - QUESTION ONE: Objective function and constraints given; solution $x=25$, $y=50$ with minimum cost 1750000. - QUESTION TWO: a) Taylor polynomials for $\ln x$ at 1: $p_0=0$, $p_1=x-1$, $p_2=x-1 - (x-1)^2/2$, $p_3=x -1 - (x-1)^2/2 + (x-1)^3/3$ b) $\sin(\pi/6 + h) = \frac{1}{2} + \frac{\sqrt{3}}{2} h - \frac{h^2}{4} - \frac{\sqrt{3}}{12} h^3 + \frac{h^4}{48}$ c) $\cos^2 2x = 1 - 4x^2 + \frac{16}{3}x^4 - \frac{128}{45} x^6$