Problem One Call the radius of the second can = r Call the height of the second can = h
Then the radius of the first can = 1/3 r The height of the first can = 3*h
A1 / A2 = (2*pi*(1/3r)*(3h)] / [2*pi * r * h]
Here's what will cancel. The twos on the right will cancel. The 3 and 1/3 will multiply to one. The 2 r's will cancel. The h's will cancel. Finally, the pis will cancel
Result A1 / A2 = 1/1 The labels will be shaped differently, but they will occupy the same area.
Problem Two It seems like the writer of the problem put some lids on the new solid that were not implied by the question.
If I understand the problem correctly, looking at it from the top you are sweeping out a circle for the lid on top and bottom, plus the center core of the cylinder.
One lid would be pi r^2 = pi w^2 and so 2 of them would be 2 pi w^2 The region between the lids would be 2 pi r h for the surface area which is 2pi w h
Put the 2 regions together and you get Area = 2 pi w^2 + 2 pi w h
Using the Law of Diminishing Marginal Utility, Jane would not be better off buying one more fry, since the marginal utility she obtains from the fries are lower when compared to the chicken sandwich; therefore, she would be better off ordering another chicken sandwich
