A microwaveable cup-of-soup package needs to be constructed in the shape of cylinder to hold 300 cubic centimeters of soup.

Question

A microwaveable cup-of-soup package needs to be constructed in the shape of cylinder to hold 300 cubic centimeters of soup. The sides and bottom of the container will be made of styrofoam costing 0.02 cents per square centimeter. The top will be made of glued paper, costing 0.08 cents per square centimeter.

Find the dimensions for the package that will minimize production cost.

Helpful information:h :

height of cylinder, r : radius of cylinder

Volume of a cylinder: 

Area of the sides: 

Area of the top/bottom: 

I have tried doing this problem a couple times with no luck. I followed this format (numbers are different than this one)

The cost of packaging, C=2πrh*0.02+ πr^2*0.02+ πr^2*0.05 subject to the constraint πr^2h=600

for this problem it would be C= 2pirh*02+pir^2*.02+pir^2h*.08 subject to cnstraint pir^2h=300

C=πr(0.04h+.07r) and the constraint implies h=600/ πr^2

for this problem: pir(.04h+.08r) contraint h=300/pir^2

 

So C=πr(24/πr^2+.07r)=24/r+.07πr^2

this problem: C=pir(12/pir^2+.07r)=12/r+.08pir^2

 

C'=-24/r^2+0.14πr=0

this problem: C'=-12/r^2+.16pir=0

r^3=24/0.14π r=3.79 cm (3 sig fig)

this problem: r^3=12/.16pi, r=2.879 which is WRONG

h=600/πr^2=13.3 cm (3 sig fig)

this problem: 300/pir^2=11.5176 which is also WRONG

C=π*3.79*(0.04*13.3+.07*3.79)=9.48cents

this problem: pi*2.879(.04*11.5176+.08*2.879)=6.25007 which is also WRONG

PLEASE HELP!!!

 

Answer 1

The .08 term in your cost equation should be πr²(.08) - just the top.

The final cost equation as you wrote it should be πr(.04h + .1r). h(r) is correct

C = 24/r + .1πr²

C' =-24/r² +.2πr

Make C' = 0 to find extremum

24 = .2πr³ r = 3.37 check C'' > 0 (minimum)

I don't know if that's enough to fix it completely, but that's what I saw.