Absolutely! Here’s a **tabular comparison** summarizing the key data, steps, and results for **Area A vs. Area B**: --- ### Tabular Comparison: Disposal Site Alternatives | Criteria | **Area A** | **Area B** | |----------------------------------|---------------------------------------------|--------------------------------------------| | **Total Capacity (million m³)** | 16 | 14 | | **Special Requirement** | Bottom 2 million m³ inert only | None | | **Inert Fill Required** | 2 million m³ | 0 | | **Inert Fill Cost (\$/m³)** | \$9.40 | — | | **Extra Haul Distance** | 0 | Weighted avg: 2.6 miles | | **Truck Speed (mph)** | 15 | 15 | | **Truck + Crew (\$/hr)** | \$140 | \$140 | | **Truck Capacity (m³)** | 20 | 20 | #### Stepwise Cost Calculation | Calculation Step | **Area A** | **Area B** | |---------------------------------|---------------------------------------------|--------------------------------------------| | **Step 1:** Special fill vol. | 2 × 10⁶ m³ × \$9.40 = **\$18.8 million** | N/A | | **Step 2:** Extra haul dist. | 0 | (0.6×5) + (0.2×–2) + (0.2×0) = **2.6 mi** | | **Step 3:** Extra haul time | N/A | 2.6 mi ÷ 15 mph = **0.1733 hr/trip** | | **Step 4:** Extra haul cost/trip| N/A | 0.1733 hr × \$140 = **\$24.27/trip** | | **Step 5:** Cost per m³ | N/A | \$24.27 ÷ 20 m³ = **\$1.213/m³** | | **Step 6:** Total extra haul cost| N/A | 14×10⁶ m³ × \$1.213 = **\$16.982 million** | | **Step 7:** **Total Cost** | **\$18.8 million** | **\$16.98 million** | --- ### **Final Recommendation** \[ \boxed{ \text{Area B is recommended since its total cost (\$16.98 million) is lower than Area A (\$18.8 million).} } \] --- Let me know if you want a more detailed breakdown or visual (e.g., bar chart)!

Certainly! Here's a concise, step-by-step solution with clear explanations, formulas, and details, followed by a tabular comparison for clarity. --- ### **Problem Restatement** A city has two rubbish disposal options: - **Area A**: - Capacity = 16 million m³ - Bottom 2 million m³ restricted to inert material only - Cost of inert material = \$9.40/m³ - **Area B**: - Capacity = 14 million m³ - No restrictions; usable for all rubbish - Additional haul distance distribution: - 60% of waste requires +5 miles - 20% requires –2 miles - 20% requires no extra distance (same as Area A) **Given Data:** - Truck speed = 15 mph - Truck + crew cost = \$140/hour - Truck capacity = 20 m³ --- ### **Step 1. Calculate Cost for Area A** The inert fill applies only to the bottom 2 million m³: \[ \text{Cost}_A = 2 \times 10^6 \, \text{m}^3 \times \$9.40/\text{m}^3 \] \[ \text{Cost}_A = \$18.8 \text{ million} \] --- ### **Step 2. Determine Weighted Average Extra Haul Distance for Area B** Calculate the average additional miles based on distribution: \[ \text{Average extra miles} = (0.60)(+5) + (0.20)(-2) + (0.20)(0) \] \[ = 3 - 0.4 + 0 = 2.6 \text{ miles} \] --- ### **Step 3. Compute Cost per Trip Due to Extra Haul** Time for extra distance: \[ t = \frac{2.6 \, \text{miles}}{15 \, \text{mph}} = 0.1733 \text{ hours} \] Cost per trip: \[ C_{\text{trip}} = 0.1733 \times \$140 = \$24.27 \] Cost per cubic meter: \[ C_{m^3} = \frac{\$24.27}{20 \, \text{m}^3} = \$1.213 \text{ per m}^3 \] --- ### **Step 4. Calculate Total Cost for Area B** Total waste volume: \[ 14 \times 10^6 \, \text{m}^3 \] Total haul cost: \[ \text{Cost}_B = 14 \times 10^6 \times \$1.213 = \$16.982 \text{ million} \] --- ### **Step 5. Final Comparison and Decision** | Criterion | Area A | Area B | |------------|---------|---------| | Total inert fill cost | \$18.8 million | N/A | | Extra haul cost | N/A | \$16.98 million | | **Total cost** | **\$18.8 million** | **\$16.98 million** | Since **Area B's total cost is lower**, **Area B is the preferred site**. --- ### **Summary:** - **Area A total cost:** \$18.8 million - **Area B total cost:** \$16.98 million **Final answer:** \[ \boxed{ \text{Choose Area B for the lower total disposal cost.} } \] --- ### **Tabular Comparison:** | **Criteria** | **Area A** | **Area B** | |-------------------------------------|----------------------------------------------|-------------------------------------------| | Capacity (million m³) | 16 | 14 | | Inert fill required (million m³) | 2 | 0 | | Inert fill cost (\$/m³) | \$9.40 | — | | Extra haul distance | 0 | Weighted average: 2.6 miles | | Truck speed | 15 mph | 15 mph | | Truck + crew cost (\$/hr) | \$140 | \$140 | | Truck capacity (m³) | 20 | 20 | | Inert fill total cost | \$18.8 million | — | | Extra haul cost | — | \$16.98 million | | **Total cost** | **\$18.8 million** | **\$16.98 million** | --- Let me know if you'd like further details or visual aids!

Let's carefully analyze the function: \[ E(x) = 3|2 - x| - |x - 5| - |x - 1| \] by breaking it at the critical points \(x=1, 2, 5\). These points are where the absolute value expressions change their expressions. --- ### 1. For \(x \leq 1\): - \(|2 - x| = 2 - x\) - \(|x - 5| = 5 - x\) - \(|x - 1| = 1 - x\) So, \[ E(x) = 3(2 - x) - (5 - x) - (1 - x) \] Simplify: \[ E(x) = 6 - 3x - 5 + x - 1 + x = (6 - 5 - 1) + (-3x + x + x) \] \[ E(x) = 0 + (-3x + 2x) = -x \] *Inequality for \(E(x) \geq 0\):* \[ -x \geq 0 \Rightarrow x \leq 0 \] **Solution in this interval:** \[ x \leq 0 \] Since we're in the region \(x \leq 1\), the solution is: \[ (-\infty, 0] \] --- ### 2. For \(1 \leq x \leq 2\): - \(|2 - x| = 2 - x\) - \(|x - 5| = 5 - x\) - \(|x - 1| = x - 1\) Evaluate: \[ E(x) = 3(2 - x) - (5 - x) - (x - 1) \] Simplify: \[ E(x) = 6 - 3x - 5 + x - x + 1 \] Combine like terms: \[ (6 - 5 + 1) + (-3x + x - x) = 2 + (-3x) \] Thus, \[ E(x) = 2 - 3x \] *Inequality:* \[ 2 - 3x \geq 0 \Rightarrow 3x \leq 2 \Rightarrow x \leq \frac{2}{3} \] Since the interval is \(x \in [1, 2]\), and \(\frac{2}{3} \approx 0.666\), which is **less** than 1, **there are no solutions** in this interval. --- ### 3. For \(2 \leq x \leq 5\): - \(|2 - x| = x - 2\) - \(|x - 5| = 5 - x\) - \(|x - 1| = x - 1\) Evaluate: \[ E(x) = 3(x - 2) - (5 - x) - (x - 1) \] Simplify: \[ 3x - 6 - 5 + x - x + 1 \] Combine: \[ (3x + x - x) + (-6 - 5 + 1) = 3x + (-10) = 3x - 10 \] *Inequality:* \[ 3x - 10 \geq 0 \Rightarrow 3x \geq 10 \Rightarrow x \geq \frac{10}{3} \] Interval is \([2, 5]\). The solution within this interval: \[ x \geq \frac{10}{3} \approx 3.333 \] So, solutions in this interval are: \[ \left[\frac{10}{3}, 5\right] \] --- ### 4. For \(x \geq 5\): - \(|2 - x| = x - 2\) - \(|x - 5| = x - 5\) - \(|x - 1| = x - 1\) Evaluate: \[ E(x) = 3(x - 2) - (x - 5) - (x - 1) \] Simplify: \[ 3x - 6 - x + 5 - x + 1 \] Combine: \[ (3x - x - x) + (-6 + 5 + 1) = (1x) + 0 = x \] *Inequality:* \[ x \geq 0 \] This is always true in \(x \geq 5\). So, the entire interval: \[ [5, \infty) \] --- ### **Final solution set:** Putting all pieces together: - From step 1: \((-\infty, 0]\) - From step 3: \([\frac{10}{3}, 5]\) - From step 4: \([5, \infty)\) Note that \(\left[\frac{10}{3}, 5\right]\) and \([5, \infty)\) connect at \(x=5\), so the union is: \[ (-\infty, 0] \cup \left[\frac{10}{3}, \infty\right) \] --- ## **Answer:** \[ \boxed{ (-\infty, 0] \cup \left[\frac{10}{3}, \infty\right) } \] This is the set of all \(x\) satisfying \(E(x) \geq 0\).

Let's analyze the inequality: \[ |x - 1| + (x - 2)^2 - 5 \geq 0 \] --- ### Step 1: Break at critical points for \(|x - 1|\): - \(|x - 1|\) changes at \(x = 1\). --- ### Step 2: Consider the two regions: #### Region 1: \(x \geq 1\) - \(|x - 1| = x - 1\) So, the inequality becomes: \[ x - 1 + (x - 2)^2 - 5 \geq 0 \] Simplify: \[ x - 1 + (x - 2)^2 - 5 \geq 0 \] \[ (x - 2)^2 + x - 6 \geq 0 \] --- #### Region 2: \(x < 1\) - \(|x - 1| = 1 - x\) The inequality becomes: \[ 1 - x + (x - 2)^2 - 5 \geq 0 \] \[ (x - 2)^2 - x - 4 \geq 0 \] --- ### Step 3: Expand \((x - 2)^2 = x^2 - 4x + 4\) --- ### Region 1: \(x \geq 1\) Inequality: \[ x^2 - 4x + 4 + x - 6 \geq 0 \] Simplify: \[ x^2 - 3x - 2 \geq 0 \] Factor: \[ x^2 - 3x - 2 = (x - 2)(x + 1) \] Solve: \[ (x - 2)(x + 1) \geq 0 \] Critical points at \(x = -1, 2\). - Sign analysis: \[ \begin{cases} x < -1: & (x - 2)<0, \quad (x + 1)<0 \Rightarrow \text{product} > 0 \\ -1 < x < 2: & (x - 2)<0, \quad (x + 1)>0 \Rightarrow \text{product}<0 \\ x > 2: & (x - 2)>0, \quad (x + 1)>0 \Rightarrow \text{product}>0 \end{cases} \] Since we're in \(x \geq 1\), the relevant parts are: - From \(x \geq 1\), the solution where the inequality is \(\geq 0\): \[ x \leq -1 \quad \text{(not in } x \geq 1) \quad \text{or} \quad x \geq 2 \] Thus, in \(x \geq 1\): \[ x \geq 2 \] --- ### Region 2: \(x < 1\) Inequality: \[ x^2 - 4x + 4 - x - 4 \geq 0 \] \[ x^2 - 5x \geq 0 \] Factor: \[ x(x - 5) \geq 0 \] Critical points at \(x = 0, 5\). - Sign analysis: \[ \begin{cases} x < 0: & x<0, \quad x - 5<0, \quad \text{product} > 0 \\ 0 < x < 5: & x>0, x-5<0, \quad \text{product}<0 \\ x > 5: & x>0, x-5>0, \quad \text{product}>0 \end{cases} \] Since \(x<1\), the relevant part: \[ x \leq 0 \] because for \(x<1\), the inequality holds when \(x \leq 0\). --- ### **Final solution set:** - From \(x < 1\): \[ x \leq 0 \] - From \(x \geq 1\): \[ x \geq 2 \] Note that at \(x=1\), check: \[ |1 - 1| + (1 - 2)^2 - 5 = 0 + 1 - 5 = -4 < 0 \] So, \(x=1\) is **not** included. At \(x=2\): \[ |2 - 1| + (2 - 2)^2 - 5 = 1 + 0 - 5 = -4 < 0 \] So, \(x=2\) is **not** included either. --- ### **Final answer:** \[ \boxed{ (-\infty, 0] \cup [2, \infty) } \] --- **Summary:** - The inequality holds for \(x \leq 0\), - and for \(x \geq 2\), - but not at the boundary points \(x=1, 2\), since at \(x=1\), the expression is negative, and at \(x=2\), it's negative. **Therefore, the solution set is:** \[ \boxed{ (-\infty, 0] \cup [2, \infty) } \]

Let's analyze the inequality: \[ |x - 1| + (x - 2)^2 - 5 > 0 \] by breaking it at the critical point \(x=1\), where the absolute value expression changes. --- ### **Case 1: \(x \geq 1\)** - \(|x - 1| = x - 1\) Substitute: \[ (x - 1) + (x - 2)^2 - 5 > 0 \] Simplify: \[ (x - 1) + (x - 2)^2 - 5 > 0 \] Expand \((x - 2)^2\): \[ x^2 - 4x + 4 \] Now the inequality: \[ x - 1 + x^2 - 4x + 4 - 5 > 0 \] Combine like terms: \[ x^2 - 3x + ( -1 + 4 - 5) > 0 \] \[ x^2 - 3x - 2 > 0 \] --- ### **Solve the quadratic inequality:** \[ x^2 - 3x - 2 > 0 \] Factor or use quadratic formula: \[ x = \frac{3 \pm \sqrt{9 - 4 \times 1 \times (-2)}}{2} = \frac{3 \pm \sqrt{9 + 8}}{2} = \frac{3 \pm \sqrt{17}}{2} \] Since the parabola opens upward, the inequality: \[ x^2 - 3x - 2 > 0 \] holds where: \[ x < \frac{3 - \sqrt{17}}{2} \quad \text{or} \quad x > \frac{3 + \sqrt{17}}{2} \] In **Case 1**, \(x \geq 1\), so the relevant solution is: \[ x > \frac{3 + \sqrt{17}}{2} \] --- ### **Case 2: \(x < 1\)** - \(|x - 1| = 1 - x\) Substitute: \[ (1 - x) + (x - 2)^2 - 5 > 0 \] Again, expand \((x - 2)^2\): \[ x^2 - 4x + 4 \] The inequality: \[ 1 - x + x^2 - 4x + 4 - 5 > 0 \] Simplify: \[ x^2 - 5x + (1 + 4 - 5) > 0 \] \[ x^2 - 5x > 0 \] Factor: \[ x(x - 5) > 0 \] This inequality holds when: - \(x < 0\), or - \(x > 5\) Since we're in the region \(x < 1\), the relevant part is: \[ x < 0 \] --- ### **Final solution:** - For \(x < 1\): \[ x < 0 \] - For \(x \geq 1\): \[ x > \frac{3 + \sqrt{17}}{2} \] At the boundary points: - At \(x=1\): \[ |1 - 1| + (1 - 2)^2 - 5 = 0 + 1 - 5 = -4 < 0 \] so \(x=1\) is **not** included. - At \(x = \frac{3 + \sqrt{17}}{2}\): \[ x^2 - 3x - 2 = 0 \] the expression equals zero, so it's not strictly greater than zero; thus, **excluded**. --- ### **Answer:** \[ \boxed{ (-\infty, 0) \cup \left( \frac{3 + \sqrt{17}}{2}, \infty \right) } \] This is the set where the inequality holds strictly.

Let's analyze the wave equation: \[ \frac{\partial^2 h}{\partial t^2} - c^2 \frac{\partial^2 h}{\partial x^2} = 0 \] using the method of characteristic coordinates. --- ### Step 1: Introduce characteristic coordinates Define: \[ \xi = x - c t, \quad \eta = x + c t \] --- ### Step 2: Compute derivatives using the chain rule The derivatives with respect to \(x\) and \(t\) transform as follows: \[ \frac{\partial}{\partial x} = \frac{\partial \xi}{\partial x} \frac{\partial}{\partial \xi} + \frac{\partial \eta}{\partial x} \frac{\partial}{\partial \eta} = \frac{\partial}{\partial \xi} + \frac{\partial}{\partial \eta} \] \[ \frac{\partial}{\partial t} = \frac{\partial \xi}{\partial t} \frac{\partial}{\partial \xi} + \frac{\partial \eta}{\partial t} \frac{\partial}{\partial \eta} = -c \frac{\partial}{\partial \xi} + c \frac{\partial}{\partial \eta} \] --- ### Step 3: Compute second derivatives Calculate \(\frac{\partial^2}{\partial t^2}\): \[ \frac{\partial^2}{\partial t^2} = \frac{\partial}{\partial t} \left( -c \frac{\partial}{\partial \xi} + c \frac{\partial}{\partial \eta} \right) \] which yields: \[ \frac{\partial^2}{\partial t^2} = c^2 \frac{\partial^2}{\partial \xi^2} - 2 c^2 \frac{\partial^2}{\partial \xi \partial \eta} + c^2 \frac{\partial^2}{\partial \eta^2} \] Similarly, for \(\frac{\partial^2}{\partial x^2}\): \[ \frac{\partial^2}{\partial x^2} = \left( \frac{\partial}{\partial \xi} + \frac{\partial}{\partial \eta} \right)^2 = \frac{\partial^2}{\partial \xi^2} + 2 \frac{\partial^2}{\partial \xi \partial \eta} + \frac{\partial^2}{\partial \eta^2} \] --- ### Step 4: Form the PDE in characteristic coordinates Substitute into the wave equation: \[ \left( c^2 \frac{\partial^2}{\partial \xi^2} - 2 c^2 \frac{\partial^2}{\partial \xi \partial \eta} + c^2 \frac{\partial^2}{\partial \eta^2} \right) - c^2 \left( \frac{\partial^2}{\partial \xi^2} + 2 \frac{\partial^2}{\partial \xi \partial \eta} + \frac{\partial^2}{\partial \eta^2} \right) = 0 \] Simplify: \[ c^2 \frac{\partial^2}{\partial \xi^2} - 2 c^2 \frac{\partial^2}{\partial \xi \partial \eta} + c^2 \frac{\partial^2}{\partial \eta^2} - c^2 \frac{\partial^2}{\partial \xi^2} - 2 c^2 \frac{\partial^2}{\partial \xi \partial \eta} - c^2 \frac{\partial^2}{\partial \eta^2} = 0 \] which reduces to: \[ -4 c^2 \frac{\partial^2}{\partial \xi \partial \eta} = 0 \] --- ### **Step 5: General solution in characteristic form** Divide both sides by \(-4 c^2\): \[ \frac{\partial^2 h}{\partial \xi \partial \eta} = 0 \] This partial differential equation implies that \(h(\xi, \eta)\) can be written as the sum of functions, each depending on only one variable: \[ h(\xi, \eta) = F(\xi) + G(\eta) \] or, equivalently: \[ h(x,t) = F(x - c t) + G(x + c t) \] --- ### **Step 6: Characteristic velocities** The characteristic curves along which the disturbances travel are given by: \[ x - c t = \text{constant} \quad \Rightarrow \quad \frac{dx}{dt} = c \] \[ x + c t = \text{constant} \quad \Rightarrow \quad \frac{dx}{dt} = -c \] --- ## **Final result:** \[ \boxed{ \text{The characteristic velocities are } \frac{dx}{dt} = +c \quad \text{and} \quad \frac{dx}{dt} = -c } \] This indicates that waves propagate along the characteristic lines \(x = ct + \text{constant}\) and \(x = -ct + \text{constant}\), moving in opposite directions with speed \(c\).

To estimate the annual interest paid on the \$1 trillion student loan debt at an average interest rate of 6%, we can use the simple interest formula: \[ \text{Interest} = \text{Principal} \times \text{Interest Rate} \] **Step 1: Convert the interest rate to decimal form** \[ 6\% = 0.06 \] **Step 2: Calculate the interest** \[ \text{Interest} = \$1,000,000,000,000 \times 0.06 \] **Step 3: Perform the multiplication** \[ \$1,000,000,000,000 \times 0.06 = \$60,000,000,000 \] **Final Answer:** \[ \boxed{\$6 \times 10^{10}} \] **Therefore, approximately \$6 \times 10^{10} \) (or 60 billion dollars) is paid in interest on student loans each year.**

Let's go through the detailed calculations step-by-step: --- ### **Given Data:** - Ultimate moment capacity: \( M_u = 600\, \text{kN·m} = 600 \times 10^3\, \text{N·m} \) - Concrete compressive strength: \( f'_c = 30\, \text{MPa} = 30\, \text{N/mm}^2 \) - Steel yield strength: \( f_y = 400\, \text{MPa} = 400\, \text{N/mm}^2 \) - Width of the section: \( b = 350\, \text{mm} \) - Total height of section: \( h = 650\, \text{mm} \) - Cover: \( 40\, \text{mm} \) - Stirrup diameter: \( 10\, \text{mm} \) - Tension reinforcement bar diameter: \( 28\, \text{mm} \) - Strength reduction factor: \( \phi = 0.9 \) --- ### **1. Effective Depth \( d \)** Calculate effective depth: \[ d = h - \text{cover} - \frac{\phi_{st}}{2} - \frac{\phi_{bar}}{2} \] where: - \(\phi_{st} = 10\, \text{mm}\) (stirrup diameter), - \(\phi_{bar} = 28\, \text{mm}\). Substitute: \[ d = 650 - 40 - \frac{10}{2} - \frac{28}{2} = 650 - 40 - 5 - 14 = 591\, \text{mm} \] --- ### **2. Nominal Moment \( M_n \)** \[ M_n = \frac{M_u}{\phi} = \frac{600 \times 10^3\, \text{N·m}}{0.9} \] Convert N·m to N·mm: \[ 1\, \text{N·m} = 10^3\, \text{N·mm} \] So: \[ M_u = 600 \times 10^3 \times 10^3 = 6.0 \times 10^8\, \text{N·mm} \] Calculate: \[ M_n = \frac{6.0 \times 10^8}{0.9} = 6.6667 \times 10^8\, \text{N·mm} \] --- ### **3. Steel Area \( A_s \) Calculation** Using the standard flexural formula: \[ M_n = A_s f_y \left( d - \frac{a}{2} \right) \] where: \[ a = \frac{A_s f_y}{0.85 f'_c b} \] Substituting \(a\) into the moment equation gives a quadratic in \(A_s\): \[ M_n = A_s f_y \left( d - \frac{A_s f_y}{2 \times 0.85 f'_c b} \right) \] which simplifies to: \[ \frac{f_y^2}{2 \times 0.85 f'_c b} A_s^2 - f_y d A_s + M_n = 0 \] --- ### **4. Coefficients for quadratic:** Calculate: \[ a_1 = \frac{f_y^2}{2 \times 0.85 \times f'_c \times b} \] Plugging in values: \[ a_1 = \frac{(400)^2}{2 \times 0.85 \times 30 \times 350} \] Calculate numerator: \[ (400)^2 = 160,000 \] Calculate denominator: \[ 2 \times 0.85 \times 30 \times 350 = 2 \times 0.85 \times 10,500 = 2 \times 8,925 = 17,850 \] Thus: \[ a_1 = \frac{160,000}{17,850} \approx 8.9636 \] Calculate: \[ b_1 = f_y \times d = 400 \times 591 = 236,400\, \text{N} \] and \[ M_n = 666,666,666.7\, \text{N·mm} \] --- ### **5. Solve quadratic equation:** Quadratic form: \[ a_1 A_s^2 - b_1 A_s + M_n = 0 \] Discriminant: \[ \Delta = b_1^2 - 4 a_1 M_n \] Calculate: \[ b_1^2 = (236,400)^2 \approx 5.59 \times 10^{10} \] \[ 4 a_1 M_n = 4 \times 8.9636 \times 666,666,666.7 \approx 4 \times 8.9636 \times 6.6667 \times 10^8 \] Calculate: \[ 8.9636 \times 6.6667 \approx 59.757 \] Thus: \[ 4 \times 59.757 \times 10^8 = 239.028 \times 10^8 = 2.39028 \times 10^{10} \] Now, the discriminant: \[ \Delta \approx 5.59 \times 10^{10} - 2.39 \times 10^{10} = 3.20 \times 10^{10} \] Square root: \[ \sqrt{\Delta} \approx \sqrt{3.20 \times 10^{10}} \approx 178,835 \] Calculate roots: \[ A_s = \frac{b_1 \pm \sqrt{\Delta}}{2 a_1} \] Pick the smaller root: \[ A_s = \frac{236,400 - 178,835}{2 \times 8.9636} = \frac{57,565}{17.9272} \approx 3,211\, \text{mm}^2 \] --- ### **6. Bar size and number:** Area of one \(\varnothing 28\) bar: \[ A_{28} = \frac{\pi}{4} (28)^2 \approx 615.8\, \text{mm}^2 \] Number of bars: \[ n = \frac{A_s}{A_{28}} = \frac{3211}{615.8} \approx 5.21 \] Round up: \[ n=6\, \text{bars} \] Arrange as 3+3 layers, total steel area: \[ 6 \times 615.8 = 3,695\, \text{mm}^2 \] which exceeds the calculated \(A_s\), ensuring safety. --- ### **Final answers:** - **Required steel area:** approximately **3211 mm\(^2\)** - **Number of \(\varnothing 28\) bars:** **6** bars arranged as 3+3 layers. This completes the design calculation based on the given data.

Let's analyze and solve the differential equation: \[ x y' + 4 y = y^6, \quad y(1) = 2 \] --- ### Step 1: Rewrite in standard form Divide both sides by \(x\) (assuming \(x \neq 0\)): \[ y' + \frac{4}{x} y = \frac{1}{x} y^6 \] --- ### Step 2: Recognize the type of ODE This is a Bernoulli differential equation: \[ y' + P(x) y = Q(x) y^n \] where: - \(P(x) = \frac{4}{x}\), - \(Q(x) = \frac{1}{x}\), - \(n=6\). --- ### Step 3: Use substitution for Bernoulli equation Set: \[ z = y^{1-n} = y^{1 - 6} = y^{-5} \] Differentiate: \[ z = y^{-5} \Rightarrow z' = -5 y^{-6} y' \] Solve for \(y'\): \[ y' = -\frac{1}{5} y^{6} z' \] --- ### Step 4: Rewrite the original in terms of \(z\) Substitute into the differential equation: \[ -\frac{1}{5} y^{6} z' + \frac{4}{x} y = \frac{1}{x} y^6 \] Divide through by \(y^6\): \[ -\frac{1}{5} z' + \frac{4}{x} y^{1-6} = \frac{1}{x} \] Recall: \[ z = y^{-5} \Rightarrow y^{1-6} = y^{-5} = z \] Thus: \[ -\frac{1}{5} z' + \frac{4}{x} z = \frac{1}{x} \] Multiply through by \(-5\): \[ z' - \frac{20}{x} z = -\frac{5}{x} \] --- ### Step 5: Solve the linear ODE in \(z\) Standard form: \[ z' + P(x) z = Q(x) \] where: \[ P(x) = -\frac{20}{x}, \quad Q(x) = -\frac{5}{x} \] Find the integrating factor: \[ \mu(x) = e^{\int P(x) dx} = e^{\int -\frac{20}{x} dx} = e^{-20 \ln x} = x^{-20} \] --- ### Step 6: Multiply through by the integrating factor \[ x^{-20} z' + x^{-20} P(x) z = x^{-20} Q(x) \] which simplifies to: \[ \frac{d}{dx} \left( x^{-20} z \right) = -\frac{5}{x} x^{-20} = -5 x^{-21} \] --- ### Step 7: Integrate both sides \[ x^{-20} z = \int -5 x^{-21} dx + C \] Integrate: \[ \int x^{-21} dx = \frac{x^{-20}}{-20} \] So: \[ x^{-20} z = -5 \times \frac{x^{-20}}{-20} + C = \frac{5}{20} x^{-20} + C = \frac{1}{4} x^{-20} + C \] Multiply both sides by \(x^{20}\): \[ z = \frac{1}{4} + C x^{20} \] --- ### Step 8: Recall \(z = y^{-5}\) \[ y^{-5} = \frac{1}{4} + C x^{20} \] Solve for \(y\): \[ y = \left( \frac{1}{\frac{1}{4} + C x^{20}} \right)^{1/5} \] --- ### Step 9: Use initial condition \(y(1) = 2\) At \(x=1\): \[ 2 = \left( \frac{1}{\frac{1}{4} + C \times 1^{20}} \right)^{1/5} \] Raise both sides to the 5th power: \[ 2^5 = \frac{1}{\frac{1}{4} + C} \] Calculate: \[ 32 = \frac{1}{\frac{1}{4} + C} \] Invert: \[ \frac{1}{32} = \frac{1}{4} + C \] Solve for \(C\): \[ C = \frac{1}{32} - \frac{1}{4} = \frac{1}{32} - \frac{8}{32} = -\frac{7}{32} \] --- ### **Final solution:** \[ \boxed{ y(x) = \left( \frac{1}{\frac{1}{4} - \frac{7}{32} x^{20}} \right)^{1/5} } \] or equivalently, \[ y(x) = \left( \frac{1}{\displaystyle \frac{1}{4} - \frac{7}{32} x^{20}} \right)^{1/5} \] which satisfies the initial condition \(y(1)=2\). --- **Summary:** - Solution form: \( y = \left( \frac{1}{\frac{1}{4} + C x^{20}} \right)^{1/5} \) - Constant: \( C = -\frac{7}{32} \) - Final explicit solution: \[ \boxed{ \boxed{ y(x) = \left( \frac{1}{\frac{1}{4} - \frac{7}{32} x^{20}} \right)^{1/5} } } \]