A cylindrical can, open at the top, is to hold 470cm^{3} of liquid. Find the height and radius that minimize the amount of material needed to manufacture the can. Enter your answer with rational exponents, and use pi to represent pi. Radius = ? Height = ?

Question

A cylindrical can, open at the top, is to hold

470 c m^{3}

of liquid. Find the height and radius that minimize the amount of material needed to manufacture the can. Enter your answer with rational exponents, and use pi to represent pi.
Radius

= ?

Height

= ?

Answer 1

Given:

V = 470 c m^{3}

We have to find height and radius

V = π r^{2} h = 470

h - \frac{470}{π r^{2}}

Plug this into the surface area equation

S A = π r^{2} + 2 π r h

= π r^{2} + 2 π r (\frac{470}{π r^{2}})

= π r^{2} + \frac{940}{r}

Differentiate SA and set to 0 solve for r

\frac{d S A}{d r} = 2 π r - \frac{940}{r^{2}} = 0

2 π r = \frac{940}{r^{2}}

r^{3} = \frac{940}{2} x

r^{3} = \frac{470}{π}

r = (\frac{470}{π})^{\frac{1}{3}}

= 5.30 c m

Now, we find h:

h = \frac{470}{π r^{2}}

= \frac{470}{π (5.30)^{2}}

= 5.32 c m

A cylindrical can, open at the top, is to hold 470cm^{3} of liquid. Find the height and radius that minimize the amount of material needed to manufacture the can. Enter your answer with rational exponents, and use pi to represent pi. Radius = ? Height = ?

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