Course:MATH110/Archive/2010-2011/003/Notes/Optimization/Problem 1

Question

An open-top cylinder is to be constructed from a sheet of steel. If we ask that the cylinder has a volume of 10 cubic metres, what are the dimension of the cylinder which requires the least amount of material to make?

Solution, part 1$\quad$

Before reading the solution, you should really try the problem for yourself. By should I mean that if you don't you will not gain any understanding, whatever you might believe, this is 100% guaranteed. So do yourself a favour, try the problem and look at the solution once you've gave it an authentic try. Solution, part 1$\quad$ Let us denote by $r$ the radius of the cylinder and by $h$ its height. Then the surface area of the open-top cylinder is given by the sides (which is a rectangle of height $h$ and of length $2\pi r$) and by the bottom (which is a circle of radius $r$), hence the total surface area is given by $\displaystyle {A=2\pi rh+\pi r^{2}}$ We would like to minimize this amount. We don't know how to deal with two variables at the same time, but since we require the volume of the cylinder to be 10, and since the volume of a cylinder is given by $\displaystyle {V=\pi r^{2}h}$ We obtain a relationship linking $r$ and $h$ as follow $10=\pi r^{2}h\qquad \iff \qquad h={\frac {10}{\pi r^{2}}}$ This allows us to substitute $h$ in the expression of the total area to obtain a function which is defined for positive values of $r$: $A(r)=2\pi r{\frac {10}{\pi r^{2}}}+\pi r^{2}=\pi r^{2}+{\frac {20}{r}}$ We need only to find the minimum value of that function on its domain, this is done in part 2.

Solution, part 2$\quad$

Before reading the solution, you should really try the problem for yourself. By should I mean that if you don't you will not gain any understanding, whatever you might believe, this is 100% guaranteed. So do yourself a favour, try the problem and look at the solution once you've gave it an authentic try. Solution, part 2$\quad$ To find the minimum value of the function $A(r)=\pi r^{2}+{\frac {20}{r}}\qquad {\text{for }}r>0$ We first compute its derivative $A'(r)=2\pi r-{\frac {20}{r^{2}}}$ And then solve it for zero to find its critical points $2\pi r-{\frac {20}{r^{2}}}=0\quad \iff \quad \pi r^{3}=10\quad \iff \quad r={\sqrt[{3}]{\frac {10}{\pi }}}$ We'll use the second derivative test to make sure this critical point is the minimum value of the function. $A''(r)=2\pi +{\frac {40}{r^{3}}}$ And clearly, this function is positive for any value of $r$ which is positive, so we can guarantee the critical point we found is a minimum. Substituting the exact value back to find $h$ shows that we get $\displaystyle {h=r={\sqrt[{3}]{\frac {10}{\pi }}}}\approx 1.47$ So the dimensions of the cylinder using the least amount of material has its radius equal its height.