A rectangular tank that is 2048 ft cubed with a square base and open top is to be constructed of sheet steel of a given thickness. Find the dimensions of the tank with minimum weight.

16ft by 16ft by 8ft.

Step-by-step 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.

e) none of the above because the measure of angle cba is 101°