Step-by-step explanation:
![A = {cos}^{ - 1} 0.5 \\ \: \: A= 60](https://tex.z-dn.net/?f=A%20%20%3D%20%20%7Bcos%7D%5E%7B%20-%201%7D%200.5%20%5C%5C%20%20%5C%3A%20%20%5C%3A%20%20%20A%3D%2060)
Answer: 0.0228
Step-by-step explanation:
please check photo explanation
Answer:
Side of 22 and height of 11
Step-by-step explanation:
Let s be the side of the square base and h be the height of the tank. Since the tank volume is restricted to 5324 ft cubed we have the following equation:
![V = s^2h = 5324](https://tex.z-dn.net/?f=V%20%3D%20s%5E2h%20%3D%205324)
![h = 5324 / s^2](https://tex.z-dn.net/?f=h%20%3D%205324%20%2F%20s%5E2)
As the thickness is already defined, we can minimize the weight by minimizing the surface area of the tank
Base area with open top ![s^2](https://tex.z-dn.net/?f=s%5E2)
Side area 4sh
Total surface area ![A = s^2 + 4sh](https://tex.z-dn.net/?f=A%20%3D%20s%5E2%20%2B%204sh)
We can substitute ![h = 5324 / s^2](https://tex.z-dn.net/?f=h%20%3D%205324%20%2F%20s%5E2)
![A = s^2 + 4s\frac{5324}{s^2}](https://tex.z-dn.net/?f=A%20%3D%20s%5E2%20%2B%204s%5Cfrac%7B5324%7D%7Bs%5E2%7D)
![A = s^2 + 21296/s](https://tex.z-dn.net/?f=A%20%3D%20s%5E2%20%2B%2021296%2Fs)
To find the minimum of this function, we can take the first derivative, and set it to 0
![A' = 2s - 21296/s^2 = 0](https://tex.z-dn.net/?f=A%27%20%3D%202s%20-%2021296%2Fs%5E2%20%3D%200)
![2s = 21296/s^2](https://tex.z-dn.net/?f=2s%20%3D%2021296%2Fs%5E2)
![s^3 = 10648](https://tex.z-dn.net/?f=s%5E3%20%3D%2010648)
![s = \sqrt[3]{10648} = 22](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%5B3%5D%7B10648%7D%20%3D%2022)
![h = 5324 / s^2 = 5324 / 22^2 = 11](https://tex.z-dn.net/?f=h%20%3D%205324%20%2F%20s%5E2%20%3D%205324%20%2F%2022%5E2%20%3D%2011)
I believe the answer is 75 I am not sure but I adds up because 4*.75 equals to 3 there is another way to solve this though we can do the is over of equals % over 100 then we still get three have a nice day and good luck on your assignment (: