Question 1183388
The thickness of the open-lid tank is 50 cm = {{{(1/2) m}}}. Let the interior square base have a side of length {{{ x }}} meters.  

Also let the inner height be {{{y }}} m.  
This then gives 

{{{x^2*y = 4000}}}

since it is given that the inside capacity is {{{4000 m^3}}}.

For the exterior of the tank, the square base has a side of length 
{{{ x + 2*(1/2) = x+1}}} meters, and a height of {{{y + 1/2}}} meters.

The external volume is then 

{{{V = (x+1)^2*(y+1/2) = (x+1)^2*( 4000/x^2 +1/2)}}}, after substituting for {{{y}}} from the previous equation.

==> {{{dV/dx = 2(x+1)(4000/x^2+1/2)-(x+1)^2(8000/x^3)

= (x+1)(8000/x^2+1)-(x+1)^2(8000/x^3) = ((x+1)*(x^3 - 8000))/x^3}}}, 

after simplifying the expression on the right side.

Setting {{{dV/dx = ((x+1)*(x^3 - 8000))/x^3 = 0}}}, we get {{{x = 20 m}}}.

We cannot accept {{{x = -1}}} since the domain of V is {{{x > 0}}}.

Now if  {{{x < 20}}}, {{{dV/dx < 0 }}}, and 

when {{{x > 20}}}, {{{dV/dx > 0 }}},  so by the the 1st derivative test, 

there is a local minimum at {{{x = 20}}}. Since it is the only critical 

point in the domain (0, {{{infinity}}}), the minimum at {{{x = 20}}} is also

absolute minimum. 

Therefore the inner dimensions of the tank are 20 m x 20 m x 10m. (The height is 

{{{y = 4000/x^2 = 4000/20^2 = 10 m}}}.)