A cylindrical tank with a radius of 5 feet is filled with water to a height of 10 feet. If the radius is increased by 50%, how much more water (in cubic feet) can the tank hold at the same height?
["A cylindrical tank with a radius of 5 feet filled with water to a height of 10 feet represents a common scenario in water storage calculations. This configuration holds approximately 1,225 cubic feet of water, a measurement often referenced in residential, agricultural, and industrial settings. With increasing focus on efficient infrastructure and water resource management in the US, such geometric calculations are gaining relevance. As trends shift toward optimizing space and storage capacity, understanding volume changes in cylindrical tanks becomes essential—especially when design parameters like radius are adjusted. This article explores how expanding the tank’s radius by 50%—from 5 feet to 7.5 feet—fires up its water-holding potential, grounded in straightforward arithmetic and real-world context.", "---", "Why This Tank Group Is Trending in 2024 \nAmerican households and businesses alike are reevaluating storage solutions amid rising demand for water efficiency and space optimization. The cylindrical tank design, rooted in simplicity and strength, remains popular for rainwater harvesting, backup water supply, and utility storage. With water conservation efforts gaining momentum and fluctuating municipal rates, upgrading from standard 5-foot radius tanks makes practical sense. The rise of smart home systems and data-driven maintenance tools further amplifies interest in precision calculations—like volume changes from structural adjustments—to guide smarter investments.", "---", "How Volume Changes When Radius Increases by 50% \nThe formula for the volume of a cylinder is \( V = \pi r^2 h \). Starting with a radius \( r = 5 \) feet and height \( h = 10 \) feet: \n- Original volume: \( V = \pi \ imes 5^2 \ imes 10 = 250\pi \) cubic feet \n- After increasing radius by 50%, new radius becomes \( 5 \ imes 1.5 = 7.5 \) feet", "New volume: \n\( V = \pi \ imes 7.5^2 \ imes 10 = \pi \ imes 56.25 \ imes 10 = 562.5\pi \) cubic feet", "The additional capacity is: \n\( 562.5\pi - 250\pi ="]
