Question 1189158
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A cylindrical can without a top is made to contain 75 in^3 of liquid. 
Find the dimensions that will minimize the cost of the metal to make the can.
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            The problem is to minimize the surface area of the described cylinder without the top.



<pre>
As you know, the volume of a cylinder is 

    V = {{{pi*r^2*h}}}, 

where pi = 3.14, r is the radius and h is the height.


In your case the volume is fixed:

    {{{pi*r^2*h}}} = 75 cubic inches.                          (1)


The surface area of a top-opened cylinder is 

    S = {{{2pi*r*h}}} + {{{pi*r^2}}},                               (2)

and they ask you to find minimum of (2) under the restriction (1).


You can rewrite the formula (2) in the form

    S(r) = {{{(2*pi*r^2*h)/r}}} + {{{pi*r^2}}}.                         (3)


In formula (3), replace  {{{pi*r^2*h}}}  by  75, based on (1). You will get

    S(r) = {{{(2*75)/r}}} + {{{pi*r^2}}} = {{{150/r}}} + {{{pi*r^2}}}.


The plot below shows the function S(r) = {{{150/r}}} + {{{pi*r^2}}}, and you can clearly see that it has the minimum.



    {{{graph( 330, 330, -5, 20, -50, 500,
          150/x + 3.14*x^2
)}}}


        Plot y = {{{150/r}}} + {{{3.14*r^2}}}



To find the minimum, use Calculus: differentiate the function to get

S'(r) = {{{-150/r^2}}} + {{{2*pi*r}}} = {{{(-150 + 2pi*r^3)/r^2}}}

and equate it to zero.


S'(r) = 0   leads you to equation  {{{2pi*r^3}}} = {{{150}}},   which gives 

r = {{{root(3,150/(2*pi))}}} = {{{root(3,150/(2*3.14))}}} = 2.88 inches (approximately).


<U>Answer</U>.  r = 2.88 inches, h = {{{75/(3.14*2.88^2)}}} = 2.88 inches  give the minimum of the surface area.
</pre>

Solved.