A cylindrical can with base radius of x cm has a tight-fitting lid which overlaps the can to a depth of 1/2 cm. The area of metal used for the can and lid is 88cm^2. Taking ㅠ as 22/7 and neglecting the thickness of the metal,

a) Show that x=2 when the volume of the can is maximum

b) Show that the volume is 56 4/7 cm^3

Question

Mathematics

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23 April, 15:31

A cylindrical can with base radius of x cm has a tight-fitting lid which overlaps the can to a depth of 1/2 cm. The area of metal used for the can and lid is 88cm^2. Taking ㅠ as 22/7 and neglecting the thickness of the metal,

a) Show that x=2 when the volume of the can is maximum

b) Show that the volume is 56 4/7 cm^3

+3

Answers (1)

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Hugh Reilly · Answer 1

Part a)

To begin with this question we must first find the formulae necessary to work with. We are dealing with surface area (SA) and volume of a cylinder.

SA = 2pi (r) h + 2pi (r^2)

V = pi (r^2) h

We are saying r = x, and we must solve for x when the volume of the can is a maximum. We also know that the SA is equal to 88 cm^2. The question states that the SA value given also includes a lid that overlaps 1/2 cm over the edge, so we need to add an extra 2pi (r) h where h = 1/2 to our formula. We can use the corrected formula to solve for h in terms of x. We are also told to use pi = 22/7.

SA = 2pi (x) h + 2pi (x^2) + pi (x) = 88

2pi (x) h = 88 - 2pi (x^2) - pi (x)

h = (88 - 2pi (x^2) - pi (x)) / (2pi (x))

h = 44 / (pi (x)) - x - 1/2

Now that we have a formula for h, we can insert this into the Volume formula.

V = pi (x^2) h

V = pi (x^2) (44/pi (x) - x - 1/2)

V = 44x - pi (x^3) - (pi/2) (x^2)

To determine the maximum volume, we must take the derivative of the volume and determine the values of x when dV/dx = 0.

dV/dx = 44 - 3pi (r^2) - pi (x) = 0

-3pi (r^2) - pi (x) + 44 = 0

a = - 3pi

b = - pi

c = 44

Now we can use the quadratic formula to solve for x.

x = (-b + / - sqr (b^2 - 4ac)) / 2a

x = (pi + / - sqr (pi^2 - 4 (-3pi) (44))) / (-6pi)

x = (22/7 + / - 286/7) / (-132/7)

x = - 7/3 and x = 2

Since we cannot use a negative integer, the radius must be equal to 2 cm at the maximum.

Part b)

Now that we have shown the radius is 2 cm at the maximum volume. We can solve for h and solve for the maximum volume.

h = 44 / (pi (x)) - x - 1/2

h = 44 / (pi (2)) - 2 - 1/2

h = (44 - 5pi) / (2pi)

V = pi (r^2) h

V = pi (2^2) (44 - 5pi) / (2pi)

V = 88 - 10pi

V = 88 - 10 (22/7)

V = 396/7

V = 56 4/7 cm^3

The maximum volume using the radius of 2 cm gives a volume of 56 4/7 cm^3.