Ummm... Are you ok? Please retype this lol
To solve this problem you must apply the proccedure shown below:
1. You must use the formula for calculate the surface area and the volume of the tank:
2. Solve for 
in the second equation and substitute it into the first one.Then, you have:
3. Now, derivate:
4. When you solve for 
, you obtain:
![2x^{3}-256=0\\ x=\sqrt[3]{\frac{256}{2}} \\ x=5.03](https://tex.z-dn.net/?f=%202x%5E%7B3%7D-256%3D0%5C%5C%20x%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B256%7D%7B2%7D%7D%20%20%5C%5C%20x%3D5.03%20)
5. Now, you must calculate :
Therefore, the dimensions are:
The sides of the base of the tank: 
The height of the tank: 
The numbers are 109 and 327. Hope it helps!
Yes you have the right idea. You find the derivative to get the velocity function and then determine when the velocity is zero. That turns out to be at t = 2 and t = 6. Both are correct. The final answer is correct.
The only flaw I see is that you wrote
x'(t) = 3y^3 - 24t + 36
on line 2 when it should be
x'(t) = 3t^2 - 24t + 36
Other than that it looks perfect.
Answer is in the photo. I can only upload it to a file hosting service. link below!
tinyurl.com/wpazsebu
