Answer:
16ft by 16ft by 8ft.
Explanation:
Let the total surface area of the rectangular tank be S = 2LW+2LH+2WH where;
L is the length of the box
W is the width of the box
H is the height of the box.
Since the box is openend at the top, S = lw + 2lh+ 2wh
If the base is a square base then, l = w
S = l(l) + 2wh+2wh
S = l²+4wh ............... 1
If volume = lwh
lwh = 2028 ft³
wh = 2048/l ................ 2
Substitute equation 2 into 1;
S = l²+4(2048/l)
S = l²+8192/l
dS/dl = 2l - 8192/l²
If dS/dl = 0 (since we are looking for dimensions of the tank with minimum weight.)
2l - 8192/l² = 0
2l = 8192/l²
2l³ = 8192
l³ = 8192/2
l³ = 4096
l =∛4096
l = 16 ft
Since the length is equal to the width, hence the width = 16ft (square based tank)
Given the volume V = lwh = 2048
lwh = 2048
16*16*h = 2048
256h = 2048
divide both sides by 256
256h/256 = 2048/256
h = 8ft
Hence, the dimensions of the tank with minimum weight is 16ft by 16ft by 8ft.