Really need help with this constraint/optimizing volume prob

[Rebel*and*Saint]

A large can in the shape of a cylinder is to be designed so it holds 16pi cubic inches (approx. 28 ounces). Find the values of the radius r and the height h of the can, for which the minimum amount of metal is used (minimum surface area)

Recall the volume of the cylinder is given by V = pi r^2h and the surface area by A= pi r^2 + 2pi r h

a) Use the constraint equation to write the optimizing equation in terms of one variable only.

b) Find the minimum of that function.

c) Write the solution to the problem for r and h

d) What is the minimum amount needed to build the can?

I have been really stuck on this one, and do not know where to begin. I have looked in my textbook and notes, but am still stuck. Thank you!

 

[galactus]

I'll help get you started.

The surface area of the can is \(\displaystyle \L\\2{\pi}rh+2{\pi}r^{2}\)

This accounts for the body of the can and the top and bottom.

The volume is given by \(\displaystyle \L\\{\pi}r^{2}h=16{\pi}\)

Solve the volume equation for h, sub into the surface equation, differentiate, set to 0 and solve for r.

 

[Rebel*and*Saint]

volume problem pi r^2h=16pi constraint/optimizing

Still having trouble getting started.

I have r=4 thus far

 

[stapel]

I have r=4 thus far

How did you obtain this value?

Please reply showing all of your steps. You took the equations provided by the tutor, solved the volume equation for h in terms of r, plugged this into the surface-area equation in place of h, differentiated, set equal to zero, and... then what?

Thank you.

Eliz.

 

[Rebel*and*Saint]

I did this

pi r^2h=16pi
h=4

a=pi r^2+2pi r(4)
a=r^2+2r(4)=0
r=8

a=pi 8^2+2pi(8)(4)
=64+64
=128 cubic inches

 

[stapel]

pi r^2h=16pi
h=4

How did you get that h equalled 4?

Thank you.

Eliz.

 

[galactus]

"I knew I should've taken that left toin at Albuquoique"--Bugs Bunny