Physics, asked by satamsarthak03, 8 months ago

A sheet of area 40m^2 is used to make an open tank of square base then find the dimensions of the square if its volume is at maximum ​

Answers

Answered by handsomeram16645
2

Explanation:

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

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