A closed cylindrical tank having a volume of 71.57 m3 is to be constructed. If the surface area is to be minimum, what is the required diameter of the tank? A. 4 m C. 5 m B. 5.5 m D. 4.5 m Solution Click here to… Jhun Vert Thu, 06/27/2024 - 21:17 Solution Click here to expand or collapse this section For maximum volume $H = D$ $V = \frac{1}{4}\pi D^2 H$ $71.57 = \frac{1}{4}\pi D^3$ $D = 4.5 ~ \text{m}$ Detailed Solution Click here to expand or collapse this section Volume of the tank $V = \frac{1}{4}\pi D^2 H$ $71.57 = \frac{1}{4}\pi D^2 H$ $H = \dfrac{286.28}{\pi D^2}$ Surface area of the tank $A = 2A_b + A_L$ $A = 2 \left( \frac{1}{4}\pi D^2 \right) + \pi DH$ $A = \frac{1}{2}\pi D^2 + \pi D \left( \dfrac{286.28}{\pi D^2} \right)$ $A = \frac{1}{2}\pi D^2 + \dfrac{286.28}{D}$ For the surface area to be minimum $\dfrac{dA}{dD} = \pi D - \dfrac{286.28}{D^2} = 0$ $\pi D = \dfrac{286.28}{D^2}$ $D^3 = \dfrac{286.28}{\pi}$ $D = 4.5 ~ \text{m}$ Log in or register to post comments Log in or register to post comments
Solution Click here to… Jhun Vert Thu, 06/27/2024 - 21:17 Solution Click here to expand or collapse this section For maximum volume $H = D$ $V = \frac{1}{4}\pi D^2 H$ $71.57 = \frac{1}{4}\pi D^3$ $D = 4.5 ~ \text{m}$ Detailed Solution Click here to expand or collapse this section Volume of the tank $V = \frac{1}{4}\pi D^2 H$ $71.57 = \frac{1}{4}\pi D^2 H$ $H = \dfrac{286.28}{\pi D^2}$ Surface area of the tank $A = 2A_b + A_L$ $A = 2 \left( \frac{1}{4}\pi D^2 \right) + \pi DH$ $A = \frac{1}{2}\pi D^2 + \pi D \left( \dfrac{286.28}{\pi D^2} \right)$ $A = \frac{1}{2}\pi D^2 + \dfrac{286.28}{D}$ For the surface area to be minimum $\dfrac{dA}{dD} = \pi D - \dfrac{286.28}{D^2} = 0$ $\pi D = \dfrac{286.28}{D^2}$ $D^3 = \dfrac{286.28}{\pi}$ $D = 4.5 ~ \text{m}$ Log in or register to post comments
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