Question

A sheet of area 40 m2 in used to make an open tank with a square base, then find the dimensions of the base
such that volume of this tank is maximum.​

Answer 1

Dear Tarun,

◆ Answer -

Dimensions of the base = 3.65 m × 3.65 m

● Explanation -

Given that area of open tank is 40 m2.

A = s^2 + 4sh

4sh = 40 - s^2

h = (40 - s^2) / 4s

Now volume of the open tank is given by -

V = s^2.h

V = s^2 (40 - s^2) / 4s

V = s (40 - s^2) / 4

V = 10s - s^3 / 4

Tank will have maximum volume when dV/ds = 0.

dV/ds = 10 - 3s^2 / 4

0 = 10 - 3s^2 / 4

3s^2 / 4 = 10

s^2 = 40/3

s = √(40/3)

s = 3.65 m

To calculate height of tank -

h = (40 - s^2) / 4s

h = (40 - 40/3) / 4(3.65)

h = 1.82 m

Therefore, dimensions of the base of the tank is 3.65 m × 3.65 m.

Hope this helped you.

Answer 2

Answer:

base of square = 40/3 m^2

Explanation:

A sheet of area 40 m2 in used to make an open tank with a square base, then find the dimensions of the base

such that volume of this tank is maximum.

let say base of open tank

= x^2

height of tank = h

area of tank = x^2 + 4xh = 40

4xh = 40 - x^2

h = (40 - x^2)/4x

volume of tank =.x^2h

V = x^2 (40 -x^2)/4x

V = 10x - x^3/4

dV/dx = 10 - 3x^2/4

dv/dx = 0

3x^2/4 = 10

x^2 = 40/3

this will give maximum volume.

base of square = 40/3 m^2

dimensions = 2 root(10/3) * 2 root (10/3)

h = (40 - x^2)/4x

= (40 -40/3)/4*2root(10/3)

=( 80/3)/8 root(10/3)

= root(10/3)

Volume = s^2h = (40/3) * root(10/3) m^3