Respuesta :
Answer: Tank dimensions:
s = 12 ft
h = 6 ft
Step-by-step explanation:
s = lenght side of the square base
A the total area of of the tank  ( this area is the area of the base plus  the lateral area (4) times
h  = the height of the tank
V  = 864 ft³       V  =  s² * h   ⇒  h = V/s²  ⇒ h = 864/ s²
Then we have :
A (total) = Â Base area + 4 * lateral area
Area of the base is  s²
Lateral area is  s * h  but we have 4 wall  so lateral area = 4*s*h
Then Objective function is :
A(s) = s² + 4*s * V/s²      ⇒   A(s) = s²  + 4*864/s   ⇒  A(s) = s²  + 3456/s
Taken derivative of the objective function:
A´(s) = 2s - 3456/ s²
Solving for s
2s - 3456/s²  = 0   ⇒  2*s³  - 3456 = 0  ⇒  s³  = 3456/2  ⇒ s³ =  1728
s = ∛1728    ⇒  s =  12 ft
So h = V/s²    ⇒  h = 864 / 144    ⇒  h = 6 ft
Answer:
Dimension : Â Â Â s = 12ft and h = 6ft
Objective function : Â Â Â Â [tex]\rm A(s) = s^2 + 4\times \dfrac {V}{s}[/tex]
Step-by-step explanation:
Given :
Volume (V) of Rectangular Tank = 864ft
Calculation:
Let s be the length sides of the square base.
Let h be the height of the tank.
Let A be the total area of the tank.
[tex]\rm A = (Base \;Area)+(4 \times Lateral \; Area)[/tex]
[tex]\rm Lateral\; Area = s \times h = s \times \dfrac {V}{s^2} = \dfrac {V}{s}[/tex]
[tex]\rm Base\;Area = s^2[/tex]
Than objective function is,
[tex]\rm A(s) = s^2 + 4\times \dfrac {V}{s} = s^2 + 4\times\dfrac{864}{s}[/tex]
Taken derivative of the area function,
[tex]\rm A{}'(s) = 2s - \dfrac {3456}{s^2}[/tex]
Now, for minimum surface area
[tex]\rm A{}'(s) = 0[/tex]
[tex]\rm 2s - \dfrac{3456}{s^2} = 0[/tex]
[tex]\rm 2\times s^3 = 3456[/tex]
[tex]\rm s^3 = 1728[/tex]
[tex]\rm s = \sqrt[3]{1728}[/tex]
[tex]\rm s = 12 ft[/tex]
We know that volume,
[tex]\rm V = s^2 \times h[/tex]
[tex]\rm 864 = 12^2 \times h[/tex]
[tex]\rm h = 6ft[/tex]
Therefore, length sides of the square base = 12ft and height of the tank = 6ft
For more information, refer the link given below
https://brainly.com/question/4527640?referrer=searchResults
