Optimization

Question

You want to produce a cylindrical water container with a capacity of 350mL. What dimensions will minimize the amount of material required for the container?

Answer 1

You want to minimize the surface area but have a volume of 350.

 

V = 350 = pi*r^2*h

 

Surface Area (SA) = pi*r^2+ 2*pi*r*h  

 

where r is the radius of the circular base of the cylinder and h is the height of the cylinder.

 

 

Solving the first equation for radius:

 

  350/h = pi*r^2  <---- call this equation ALPHA

 

 350/(pi*h) = r^2

 

 r = square-root( 350/(pi*h)) =  (350/(pi*h))^1/2 <--- call this equation BETA

 

Now we substitute equations ALPHA and BETA into the surface area formula

so that we get a function for the surface area in terms of only h.

Let's call it SA(h)

 

SA = pi*r^2 + 2*pi*r*h

  

    =  350/h + 2 * pi * r * h  <---- equation ALPHA

 

    = 350/h + 2 * pi * (350/(pi*h))^1/2    *h <-- equation BETA

 

   = 350/h +  2 * pi * 350^(1/2)*h/ (pi*h)^1/2

  

   = 350/h + 2 * pi * 350^(1/2)*h /  (pi^(1/2)*h^(1/2))

 

   = 350/h + 2 * (pi)^(1/2) * 350^(1/2) * h^(1/2) 

 

   = 350/h + 2*pi^(1/2) * 350^(1/2) * h^(3/2) / h

 

   = (350 + 2* (pi*350)^(1/2) * h^(3/2))/h

 

Now we have to find the extrema of this function.

We take the derivative and set it equal to zero. Then solve for h.

 

0 = SA'(h) = dSA/dh =

 

   h * [(3( square-root(350*pi)*h^(1/2) - 350 - 2*square-root(350*pi)*h^(3/2)]/h^2  <---- quotient rule

 

    =[3*h^3/2*square-root(350*pi) - 350 - 2*square-root(350*pi)*h^(3/2)]/h^2

 

   = [h^3/2 * square-root(350*pi) - 350]/h^2

 

 

Solving for h:     h^(3/2)*square-root(350*pi) - 350 = 0

 

                        h^(3/2) = square-root(350/pi)

 

                       h = (350/pi)^(1/2)*2/3 = (350/pi)^(1/3)

 

 

 

The radius is then  350 = pi*r^2*h

 

                           350 = pi*r^2 * (350/pi)^(1/3)

 

                          350 * (pi/350)^(1/3) * pi^(-1) = r^2

 

                         (350/pi)* (pi/350)^(1/3) = r^2

 

                        ( 350/pi)^(2/3) = r^2

 

                        (350/pi)^(1/3) = r

 

So the optimal radius r and height h are both (350/pi)^(1/3) 

                              = cube-root(350/pi)

 

The volume is 350 with dimensions as:

V = pi*r^2*h = pi * ((350/pi)^1/3)^2 * (350/pi)^(1/3)

 

                     = pi * (350/pi)^(2/3) * (350/pi)^(1/3)

     

                       pi * (350/pi) = 350

 

                          

 

With these dimensions, I get a surface area of 701*pi*(350/pi)^(2/3)

 

 

 

Answer 2

You need these formulas,

 

350 = πr²h

 

Perimeter = 2(2πr + h) = 4πr + 2h

 

where:

r = radius

h = height

 

You want to take the derivative of perimeter and set it equal to zero.  But first, substitute h for r.

 

Perimeter = 4πr + 2(350 / πr²)

Perimeter = 4πr + (700 / πr²)

 

 

Now you can take the derivative of perimeter and set it equal to zero.  Solve for radius to find the critical values.  If the derivative changes from negative to positive around the value of radius, then that radius will minimize material.  Once you have the radius, substitute into volume equation to find height.  Keep in mind that both radius and height must be positive values.