by Guest » Thu Feb 20, 2014 8:02 pm
I assume if possible no wastage of steel plate.
Let W equal width of tank and 2W is the length and depth is ( W - 4 )
Base is ( W x 2W )
Sides 2 off them....( 2W x (W-4) )
Ends 2 of them.... ( W x (W-4)
The 2 ends can be made from same size as sides but cut in 2 pieces.
So... need a sheet of plate ( 2W x ( 4W - 12 ) )
and the area we are told of this is 1440 m^2.
So..... 8W^2 - 24W - 1440 = 0 .....solve for W
So...... W^2 - 3W - 180 = 0
Factorises.... ( W + 12 ) ( W - 15 ) = 0
So...... W - 15 = 0 .....So ..... W = 15
So size of plate is 30 x 48 = 1440 m^2
Size of tank is....15m x 30m x 11m = 4950 m^3
So it all works out evenly and no wastage..........
...................Simple...................