Question

A rectangular tank that is 4000 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.

Answer 1

dimension is 20ft*20ft*10ft

Step-by-step explanation:

let a,a and h are length, width and height of given tank with square base.

volume = a*a*h

`4000ft^3 = a^2 *h`

`h =(4000)/(a^2)`

surface area =aa +2(ah+ah)

`= a^2 + 4ah`

`= a^2 + 4r((4000)/(a^2))`

`=a^2 + ((16000)/(a))`

for weight to minimize, surface area is to be minimum i.e.

`(dA)/(da) = 0`

`(dA)/(da) = 2a -((16000)/(a^2))`

`2a = ((16000)/(a^2))`

`a^3 = 8000`

a = 20ft

now

`(d^2A)/(dr^2) = 2+((32000)/(r^2))`

at a = 20 ft

`h =(4000)/(20^2)`

h = 10ft

hence dimension is 20ft*20ft*10ft

Answer 2

dimension of the rectangular tank = (20 ft × 20 ft × 10 ft)

Step-by-step explanation:

volume of rectangle = 4000 ft³

volume of the tank = a² × h

surface area of the tank = 4 × a × h + a²

from the volume of the tank h = 4000/a²

now surface area becomes

`S = a^2 + (16000)/(a)`

now ,

`\frac{\mathrm{d} s}{\mathrm{d} a}= 2a - (16000)/(a^2)`

`\frac{\mathrm{d} s}{\mathrm{d} a}= 0\\2a - (16000)/(a^2)=0\\a^3 = 8000\\a=20 ft`

h = 10 ft.

hence, the dimension of the rectangular tank comes out to be

(20 ft × 20 ft × 10 ft)