### QUESTION ONE: Linear Programming Problem (LPP) 1. **Problem Statement:** A petroleum company operates two refineries with associated costs and production capacities. The goal is to minimize the operating cost while meeting the oil order requirements. 2. **Define variables:** Let $x$ = number of days refinery 1 operates Let $y$ = number of days refinery 2 operates 3. **Objective Function:** Minimize total cost: $$\text{Cost} = 20000x + 25000y$$ 4. **Constraints:** Based on production requirements: - High-grade oil: $$400x + 300y \geq 25000$$ - Medium-grade oil: $$300x + 400y \geq 27000$$ - Low-grade oil: $$200x + 500y \geq 30000$$ Also, days can't be negative: $$x \geq 0, \quad y \geq 0$$ 5. **Solving the LPP:** Use the constraints to find feasible $x,y$ and minimize cost. - From the constraints, form equations and solve for $x,y$ by using substitution or elimination methods. 6. **Calculate intersection points by solving pairs of constraints:** (i) Solve $$400x + 300y = 25000$$ $$300x + 400y = 27000$$ Multiply first by 4 and second by 3: $$1600x + 1200y = 100000$$ $$900x + 1200y = 81000$$ Subtract to eliminate $y$: $$700x = 19000 \Rightarrow x = \frac{19000}{700} = 27.143$$ Then $$300(27.143) + 400y = 27000 \Rightarrow 8142.9 + 400y = 27000 \Rightarrow 400y=18857.1 \Rightarrow y=47.143$$ (ii) Solve $$300x + 400y = 27000$$ $$200x + 500y = 30000$$ Multiply first by 5 and second by 4: $$1500x + 2000y = 135000$$ $$800x + 2000y = 120000$$ Subtract: $$700x = 15000 \Rightarrow x = 21.429$$ Then $$300(21.429) + 400y = 27000 \Rightarrow 6428.57 + 400y=27000 \Rightarrow 400y=20571.43 \Rightarrow y=51.429$$ (iii) Solve $$400x + 300y = 25000$$ $$200x + 500y = 30000$$ Multiply first by 5 and second by 3: $$2000x + 1500y=125000$$ $$600x + 1500y=90000$$ Subtract: $$1400x=35000 \Rightarrow x=25$$ Then $$400(25) + 300y=25000 \Rightarrow 10000 + 300y=25000 \Rightarrow 300y=15000 \Rightarrow y=50$$ 7. **Check feasibility and costs:** - Point (27.143, 47.143): Cost $=20000(27.143)+25000(47.143) = 542,860 + 1,178,575 = 1,721,435$ - Point (21.429, 51.429): Cost $=20000(21.429)+25000(51.429) = 428,580 + 1,285,725 = 1,714,305$ - Point (25, 50): Cost $=20000(25)+25000(50)=500,000+1,250,000=1,750,000$ Minimum cost at (21.429, 51.429) days. --- ### QUESTION TWO: Taylor Polynomials and Series 1. **Problem Statement:** Find Taylor polynomials and expansions for $f(x) = \ln(x)$ at $x=1$. 2. **Taylor Polynomial formula:** $$p_n(x) = f(1) + f'(1)(x-1) + \frac{f''(1)}{2!}(x-1)^2 + \frac{f^{(3)}(1)}{3!}(x-1)^3 + ...$$ 3. **Find derivatives:** - $f(x) = \ln(x)$ - $f'(x) = \frac{1}{x}$ - $f''(x) = -\frac{1}{x^2}$ - $f^{(3)}(x) = \frac{2}{x^3}$ Evaluate at $x=1$: - $f(1) = 0$ - $f'(1) = 1$ - $f''(1) = -1$ - $f^{(3)}(1) = 2$ 4. **Write polynomials:** - $p_0 = 0$ - $p_1 = 0 + 1(x-1) = (x-1)$ - $p_2 = (x-1) - \frac{1}{2}(x-1)^2$ - $p_3 = (x-1) - \frac{1}{2}(x-1)^2 + \frac{2}{6}(x-1)^3 = (x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3$ 5. **Taylor series expansion:** Using the general term for $\ln(x)$ about $x=1$: $$\ln(x) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x-1)^n}{n}$$ Expand up to the term in $(x-1)^3$: $$ (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} $$ 6. **Power series** similarly for Taylor series as above. --- ### QUESTION THREE and QUESTION FOUR No specific data given for question three and four.