2. Optimisation Exercises

By Dan Chicos. Last modified: 17 May 2023.

Question 1

A 3 m piece of wire is cut into 2 pieces and bent around to form a square and a circle.

Find the size of the 2 lengths, correct to 2 decimal places, that will make the total area of the square and circle a minimum.

Solution

Let `x` be the length which forms the square and `y` the length which forms the circle.

`x+y=3 => y=x-3`

`x` is the perimeter of a square of side `x/4` and area

`A_{1}=(x/4)^2=x^2/16`

`y` is the circumference of a circle of radius `r`.

`C=2pi r=y => r=y/(2pi)` and area

`A_{2}=pir^2=pixxy^2/(4pi^2)=y^2/(4pi)`

But `y=3-x`, therefore

`A_{2}=(3-x)^2/(4pi)=(9-6x+x^2)/(4pi)`

The total area of the square and circle is `A=A_{1}+A_{2}`

`A=x^2/16+(9-6x+x^2)/(4pi)`

`= (pi x^2)/(16pi)+(4(9-6x+x^2))/(16pi)`

`=(pix^2+36-24x+4x^2)/(16pi)`

`=((pi+4)x^2-24x+36)/(16pi)`

`A` is a parabola concave up with its minimum turning point found by solving

`(dA)/(dx)=0`

`(dA)/dx=(2(pi+4)x-24)/(16pi)=0`

`2(pi+4)x-24=0`

`x=24/(2(pi+4))=1.68`

`y=3-x=3-1.68`

`= 1.32`

`:.` The total area `A` is minimum when

`x=1.68\ "m"` and

`y=1.32\ "m".`

Question 2

A certain cylindrical soft drink can is required to have a volume of `250" cm"^3.`

(a) Show that the height of the can is `250/(pir^2)`, where `r` is the base radius.

(b) Show that the total surface area is `S=2pi r^2+500/r.`

(c) Show that `r=5/pi^(1/3)` gives a global minimum of `S` in the domain `r > 0.`

(d) Show that to minimise the area of the can, the diameter of its base should equal its height.

Solution (a)

Let the height of the can be `h" cm".`

Then volume `V=pir^2h`

`V=pir^2h=250`

` h=250/(pir^2)`

Solution (b)

Each end of the can has area `pir^2` and the curved side has area `2pirh`.

`S=2pir^2+2pirh`

`=2pir^2+2pirxx250/(pir^2)`

`=2pir^2+500/r`, where `r>0`.

Solution (c)

Differentiating,

`(dS)/(dr)=4pir-500/r^2`

`=(4pir^3-500)/r^2`

To find stationary points, solve `(dS)/(dr)=0`.

`(4pir^3-500)/r^2=0`

`4pir^3-500=0`

`4pir^3=500`

`r^3=500/(4pi)=125/pi`

`r=(125/pi)^(1/3)=5/(pi^(1/3))`

Differentiating again,

`(dS^2)/(dr^2)=d/(dr)(4pir-500/r^2)`

`=4pi-500(-2/r^3)`

`=4pi+1000/r^3>0` for all `r>0`.

Hence

`r=5/(pi^(1/3))` gives a global minimum of `S` in the domain `r>0`.

Solution (d)

When `r=5/(pi^(1/3))`, `h=250/(pir^2)`

`=250/(pi(5/(pi^(1/3)))^2)=250/(pi(25/pi^(2/3))`

`=250/(pi^(1/3)xx25)=10/pi^(1/3)`

`=2(5/pi^(1/3))`

`=2r`

Hence the minimum surface area occurs when the diameter `d=2r` equals the height.