Calculus/Maxima & Minima in Plane & Solid Figures
Expert: Paul Klarreich - 4/24/2007
QuestionA popular size of tin can with "normal" proportions has a diameter of 7.3 cm and an altitide of 10.6cm. I have to find the radius and altitude of the can such that it has minimum surface area. I am having a little trouble telling the difference between when I am finding minimum and maximum.
AnswerQuestioner: Colleen
Category: Calculus
Subject: Maxima & Minima in Plane & Solid Figures
Question: A popular size of tin can with "normal" proportions has a diameter of 7.3 cm and an altitude of 10.6cm. I have to find the radius and altitude of the can such that it has minimum surface area. I am having a little trouble telling the difference between when I am finding minimum and maximum.
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Hi, Colleen,
There is a difficulty with the way this problem is expressed. If you are already given the diameter (twice the radius, of course) and the altitude, then the problem is over.
ON THE OTHER HAND, if the problem is expressed this way:
A can of peas (not canapes, can of peas) has diameter of 7.3 cm and an altitude of 10.6cm. Find the radius and altitude of a can of peas HAVING THE SAME VOLUME that minimizes the surface area.
In that case, you will have the volume fixed, which will allow eliminating a variable. It goes like this:
A = 2(circles for top and bottom) + side(which is a rectangle = circumference * height)
A = 2 pi r^2 + 2pi rh
Now you must eliminate a variable. Use:
V = pir^2 h, and
V = pi (7.3)^2 (10.6)
So
pi (7.3)^2 (10.6) = pi r^2h
(7.3)^2 (10.6) [whatever that is]
h = -----------------------------------------
r^2
Now the rest is routine. Substitute that for h, so you have V(r). Now get V'(r), set that equal to zero, and solve.
About that max-min stuff. You might get more than one solution, AND you should note what the endpoints of the practical interval are:
h = 0 means the can must be very wide and use a lot of metal. (max, which is bad)
r = 0 means the can must be very high and use a lot of metal. (max, which is bad)
Your solution, in between, will be the minimum.