Differentiation

Question

I have to make an open cylindrical out of 48 square feet of material. That is, this can will not have a top. What are the dimensions of the radius and height that will maximize the volume of the can . calculus work.

 

 

i got 10.3 for h 

5.2 for r

i don't think it's correct please help.

Answer 1

Want to maximize Volume.  So write the equation for Volume of the can.  

 

V = (pi)r²h     

 

Need the other information to put the equation for Volume into one variable.  It is about surface area.  The surface area of a cylynder is 2(pi)rh for the sides, and (pi)r² for the bottom of the can.  This has to = 48 square feet of material.  So this is called the secondary equation:  2(pi)rh + (pi)r² = 48   Solve this for h, then substitute what you get into the primary equation (the one you are trying to maximize, the one you will take the derivative of, the Volume equation for this problem.)

 

V=(pi)r²[48-(pi)r²]/2(pi)r

 

Now simplify this as much as possible.

 

V = [(pi)r²]/[2(pi)r]  * (48 - (pi)r²)

 

V = [r/2](48 - (pi)r²)   Distribute the (r/2) into the perentheses.

 

V = [24r] - [(pi)r³]/2    

 This is as simple as it gets, and you can now take the derivative of the Volume so as to maximize it.

 

 

dV/dr = [24] - [(pi)/2](3r²)

 

Set this = to zero and solve for r.

 

I get that r = ±4/√(pi).  I would think the radius was positive, so I would say that 

 

r = 4/sqrt(pi)  and h is found by plugging it into the equation you got for h before you substituted it into the primary equation.

 

I get that h = 4/sqrt(pi)  which is approximately  2.25676 feet  and that the radius is also approximately 2.25676 feet.