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`.
So
`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.