a. Determining the maximum steel ratioFrom the given data, we can use the equation below to determine the maximum steel ratio:fy/0.85 × f´c = 0.68 × ρ × (1 - ρ/0.85)ρ = Maximum steel ratioThe values of f´c and fy are given as 30 MPa and 414 MPa, respectively.fy = 414 MPa = 414 × 10³ kPa0.85 × f´c = 0.85 × 30 MPa = 25.5 MPa = 25.5 × 10³ kPaSubstitute the values of fy and f´c into the equation to obtain:ρ = fy / 0.85 × f´cρ = (414 × 10³ kPa) / (0.85 × 25.5 × 10³ kPa)ρ = 0.6092 ≈ 0.61Therefore, the maximum steel ratio is 0.61.b. Determining the maximum moment that the beam can supportFrom the given data, we can use the equation below to determine the maximum moment that the beam can support:Mcr = (0.36 fy / 0.85 × f´c) × bd²Mcr = (0.36 × 414 MPa) / (0.85 × 30 MPa) × (0.3 m) × (0.6 m)²Mcr = 181.35 kN mTherefore, the maximum moment that the beam can support is 181.35 kN m.c. Determining the tension steel area if the beam is to resist an ultimate moment of 650 kN m.The tension steel area, As, can be determined using the formula below:Mu = φbwd²/2 × (1 - sqrt(1 - 2Mu/φbwd²fyAs)))As = (2Mu/φbwd) / (fy(1 - sqrt(1 - 2Mu/φbwd²fy)))Given, Mu = 650 kN m, φ = 0.9, b = 300 mm, d = 600 mm, fy = 414 MPa, f´c = 30 MPaThe value of φ is given as 0.9, the equation for Mu is also given as 650 kN m, b and d are given as 300 mm and 600 mm, respectively. fy is given as 414 MPa, and f´c is given as 30 MPa.Substitute the values into the equation above to obtain:As = (2Mu/φbwd) / (fy(1 - sqrt(1 - 2Mu/φbwd²fy)))As = (2 × 650 × 10³ N m / 0.9 × 0.3 m × 0.6 m²) / (414 × 10³ N/m²(1 - sqrt(1 - 2 × 650 × 10³ N m / (0.9 × 0.3 m × 0.6 m)² × 414 × 10³ N/m²)))As = 804.79 × 10⁻⁶ m² ≈ 805 mm² (answer more than 100 words).Conclusion: The maximum steel ratio is 0.61. The maximum moment that the beam can support is 181.35 kN m. The tension steel area, As, is 805 mm² if the beam is to resist an ultimate moment of 650 kN m.