A cylindrical can is to be constructed to hold a fixed volume of liquid. The cost of the
material used for the top and bottom of the can is 3 cents per square inch, and the cost of the
material used for the curved side is 2 cents per square inch. Use calculus to derive a simple
relationship between the radius and height of the can that is the least costly to construct.
Solution;
"Volume=\u03c0r^2h"
Cost of material;
"C=3(2\u03c0r^2)+2(2\u03c0rh)=6\u03c0r^2+4\u03c0rh"
"But ;"
"h=\\frac{V}{\u03c0r^2}"
Substituting into the cost;
"C=6\u03c0r^2+4\u03c0r(\\frac{V}{\u03c0r^2})"
"C=6\u03c0r^2+4\\frac Vr"
Differentiate cost w.r.t r;
"\\frac{dC}{dr}=12\u03c0r-\\frac{4V}{r^2}=0"
We have ;
"r^3=\\frac{4V}{12\u03c0}=\\frac{V}{3\u03c0}"
"r=(\\frac{V}{3\u03c0})^{\\frac13}"
While;
"h=\\frac{V}{\u03c0r^2}=\\frac{V}{\u03c0(\\frac{V}{3\u03c0})^{\\frac13}}"
Hence;
"h=2.08(\\frac{V}\u03c0)^{\\frac13}=1.42V^{\\frac13}"
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