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the total surface area of a closed cylinder is 5000cm squared. find the dimensions of the cylinder that maximizes its volume

Sagot :

The dimensions of the cylinder that maximizes its volume are radius = 16.29 cm and height = 32.57 cm

Total surface area of a cylinder

The total surface area of a cylinder is given by A = 2πr² + 2πrh where

  • r = radius of cylinder and
  • h = height of cylinder.

Since the total surface area of the cylinder A = 5000 cm². So,

A = 2πr² + 2πrh

5000 = 2πr² + 2πrh

Making h subject of the formula, we have

h = 2500/πr - r

Volume of a cylinder

Since the volume of the cylinder V = πr²h where

  • r = radius of cylinder and
  • h = height of cylinder.

Substituting h into V, we have

V = πr²h

V = πr²(2500/πr - r)

V = 2500r - πr³

Maximizing the volume

Since we want to maximize V, we differentiate with respect to r and equate to zero.

So, dV/dr = d(2500r - πr³)/dr

dV/dr = 2500 - 3πr²

Equating dV/dr to zero, we have

dV/dr = 0

2500 - 3πr² = 0

2500 = 3πr²

Dividing both sides by 3π, we have

r² = 2500/3π

Findng the radius of the cylinder

Taking square root of both sides, we have

r = √(2500/3π)

r = 50/√(3π) cm

r = 50/√(9.4247) cm

r = 50/3.07

r = 16.29 cm

Value of radius that gives maximum volume

We need to determine if this value of r gives a maximum for V. So, we differentiate dV/dr.

So, d(dV/dr)/dr = d²V/dr²

d²V/dr² = d(2500 - 3πr²)/dr

d²V/dr² = 0 - 6πr

d²V/dr² = - 6πr

Substituting the value of r into the equation, we have

d²V/dr² = - 6πr

d²V/dr² = - 6π(50/√(3π))

d²V/dr² = - 6π(50/√(3π))

Since d²V/dr² = - 6π(50/√(3π)) < 0. V is maximum at r = 50/√(3π) cm

Finding the height of the cylinder

Substituting the value of r into h, we have

h = 2500/πr - r

h = 2500/π(16.29) - 16.29

h = 2500/51.17 - 16.29

h = 48.86 - 16.29

h = 32.57 cm

So, the dimensions of the cylinder that maximizes its volume are radius = 16.29 cm and height = 32.57 cm

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