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diamong [38]
3 years ago
14

You have been asked to design a can shaped like right circular cylinder that can hold a volume of 432π-cm3. What dimensions of t

he can (radius and height) will use the least amount of material?(Hint: The lightest can will be the can with the smallest surface area.)
Mathematics
1 answer:
rosijanka [135]3 years ago
3 0

Answer:

Height = 12cm

Radius = 6cm

Step-by-step explanation:

Given

Represent volume with v, height with h and radius with r

V = 432\pi

Required

Determine the values of h and r that uses the least amount of material

Volume is calculated as:

V = \pi r^2h\\

Substitute 432π for V

432\pi = \pi r^2h

Divide through by π

432 = r^2h

Make h the subject:

h = \frac{432}{r^2}

Surface Area (A) of a cylinder is calculated as thus:

A=2\pi rh+2\pi r^2

Substitute \frac{432}{r^2} for h in A=2\pi rh+2\pi r^2

A=2\pi r(\frac{432}{r^2})+2\pi r^2

A=2\pi (\frac{432}{r})+2\pi r^2

Factorize:

A=2\pi (\frac{432}{r} + r^2)

To minimize, we have to differentiate both sides and set A' = 0

A'=2\pi (-\frac{432}{r^2} + 2r)

Set A' = 0

0=2\pi (-\frac{432}{r^2} + 2r)

Divide through by 2\pi

0= -\frac{432}{r^2} + 2r

\frac{432}{r^2} = 2r

Cross Multiply

2r * r^2 = 432

2r^3 = 432

Divide through by 2

r^3 = 216

Take cube roots of both sides

r = \sqrt[3]{216}

r = 6

Recall that:

h = \frac{432}{r^2}

h = \frac{432}{6^2}

h = \frac{432}{36}

h = 12

Hence, the dimension that requires the least amount of material is when

Height = 12cm

Radius = 6cm

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