Question 785437
{{{W}}}= width in meters
{{{L=2W}}}= length in meters
{{{D=W-4}}}= depth in meters
 
Obviously, the entire 1440m^2 of the steel plate will be used, without wasting any, or stetching/deforming the material.
That is probably far from realistic, but this is just a math problem.
We cannot expect to work on the design, because we have not given enough information. The design is someone else's problem.
 
The total surface of the tank includes the bottom, {{{WL}}}{{{m^2}}},
and the sides, {{{2(W+L)D}}}{{{m^2}}}.
Our equation is
{{{WL+2(W+L)D=1440}}}
Since L and W came expressed in terms of W, we can substitute those expressions to get an equation in terms of W only:
{{{W(2W)+2(W+2W)(W-4)=1440}}}
{{{2W^2+2(3W)(W-4)=1440}}}
{{{2W^2+6W(W-4)=1440}}}
{{{2W^2+6W^2-24W=1440}}}
{{{2W^2+6W^2-24W-1440=0}}}
{{{8W^2-24W-1440=0}}}
Dividing both sides of the equal sign by 8, we get
{{{W^2-3W-180=0}}}
Factoring, we get
{{{(W-15)(W+12)=0}}} with solutions {{{W=15}}} and {{{W=-12}}}
We throw away {{{W=-12}}} because the width in meters cannot be a negative number.
{{{w=15}}} --> {{{l=2W=2*15=highlight(30)}}}
The length of the tank is {{{highlight(30m)}}}