Question

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

Answer 1

Answer:

Square base of side = 2 × (10)^½ meters

Height = (80/3) / (8 × (10/3)^½) meters

Explanation:

We should note that the surface area of the sheet is equal to the total surface area of the open tank formed.

Let the tank be of base sides x and height h, the surface area of the open tank is given by :

SA = x² + 4(x × h)

= x² + 4xh

We have :

x² + 4xh = 40

We make h the subject of the equation:

4xh = 40 - x²

h = (40 - x²) / 4x

Volume of the tank = base Area × h

Substituting we have :

Volume = x² × (40 - x²)/4x

Canceling the x we have : x(40 - x²)/4

V = (40x - x³)/4

At maximum volume :

dV/dx = 0

Given that the volume is a quotient we will apply quotient rule in differentiation.

Given f/g :

d/x (f/g) = {f'g - fg'} / g²

We apply this in getting the derivative of the volume :

= {4(40 - 3x²) - (40 - x³) × 0}/4²

(160 - 12x²)/16 = 0

160 - 12x² = 0

12x² = 160

x² = 160/12

x² = 40/3

x = 2 × (10/3)^½ meters

h = (40 - 40/3) / (4 × 2 × (10/3)^½

h = (80/3) / 8 × (10/3)^½ meters

Answer 2

Answer :-

→ dimensions of base = √(40/3) .

Step-by-step explanation :-

Let side of square tank be x .

And, height be y .

Then, Volume = x²y .

And, Surface area = x² + 4xy .

→ 40 = x² + 4xy .

→ y = ( 40 - x² ) / 4x .

Then, V(y) = x² ( 40 - x² )/4x.

= x( 40 - x² ) / 4 .

Now, dV/dx = ( 40 - 3x² )/4 .

And, d²V/dx² = -3x/2 = Vmax.

Therefore, dV/dx = 0 .

→ ( 40 - 3x² )/4 = 0 .

→ 40 - 3x² = 0 .

→ 3x² = 40 .

→ x² = 40/3 .

$\therefore$ x = √(40/3) m.

Hence, it is solved.