Arectangular tank with a square base, an open top, and a volume of 864 ft cubed is to be constructed of sheet steel. find the dimensions of the tank that has the minimum surface area. let s be the length of one of the sides of the square base and let a be the surface area of the tank. write the objective function.

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) =  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:   x = 3 + √11     and     x = 3 - √11

step-by-step explanation:

x² - 6x + 9 = 11

(x - 3)² = 11

x - 3 = ±√11

    x = 3 ± √11

x = 3 + √11     and     x = 3 - √11