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

Answer:

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

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

$S = a^2 + \dfrac{16000}{a}$

now ,

$\frac{\mathrm{d} s}{\mathrm{d} a}= 2a - \dfrac{16000}{a^2}$

$\frac{\mathrm{d} s}{\mathrm{d} a}= 0\\2a - \dfrac{16000}{a^2}=0\\a^3 = 8000\\a=20 ft$

h = 10 ft.

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

(20 ft × 20 ft × 10 ft)

Answer 2

Answer:

dimension is 20ft*20ft*10ft

Explanation:

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

volume = a*a*h

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

$h =\frac{4000}{a^2}$

surface area =aa +2(ah+ah)

                    $= a^2 + 4ah$

                     $= a^2 + 4r(\frac{4000}{a^2})$

                        $=a^2 + (\frac{16000}{a})$

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

$\frac{dA}{da} = 0$

$\frac{dA}{da} = 2a -(\frac{16000}{a^2})$

$2a = (\frac{16000}{a^2})$

$a^3 = 8000$

a = 20ft

now

$\frac{d^2A}{dr^2} = 2+(\frac{32000}{r^2})$

at a = 20 ft

$h =\frac{4000}{20^2}$

h = 10ft

hence dimension is 20ft*20ft*10ft