Application of Calculus:
Minima/Maxima problems
In minima/maxima problems you will be given a function or will have to formulate your own
function. You will have to, using calculus, determine the value of a variable that will optimise the
function. In order to determine the maximum or minimum you will have to determine the derivative
and set it equal to zero.
This is because any function that has a local minimum or maximum by virtue of the definition has
turning points, and as we recall the turning point is found at the point where 𝑓 ′ (𝑥) = 0.
Incredibly valuable in mastering this section is the realisation that any algebraic function can be
represented by a graph, this means that all our properties of graphs from the last three lessons are
still applicable here.
Steps to solving Minima/Maxima problems:
1. Draw a diagram if possible
2. If not given, determine an expression for the quantity that needs to be optimised, in terms
of one variable.
3. If there are 2 variables, eliminate one using the given information and simultaneous
equations
4. Write down 𝑓(𝑥) & 𝑓′(𝑥)
5. Set the derivative equal to 0 for minimum or maximum
6. Solve and answer question
1
, Example:
The functions 𝒇(𝒙) = 𝒙𝟐 − 𝟔𝒙 + 𝟓 and 𝒈(𝒙) = 𝟒𝒙 + 𝟏 are sketched below. 𝑴 is a point on 𝒈(𝒙)
and 𝑵 is a point on 𝒇(𝒙) such that 𝑵 lies directly underneath 𝑴, such that 𝑴𝑵 // to the 𝒚 −axis.
a) Find an expression for the length of 𝑴𝑵.
b) Determine the maximum length of 𝑴𝑵.
a) 𝑀𝑁 = 𝑔(𝑥) − 𝑓(𝑥) The length is the higher 𝑦-value less the lower 𝑦-value
∴ 𝑀𝑁 = 4𝑥 + 1 − (𝑥 2 − 6𝑥 + 5 )
∴ 𝑀𝑁 = 4𝑥 + 1 − 𝑥 2 + 6𝑥 − 5
∴ 𝑀𝑁 = −𝑥 2 + 10𝑥 − 4
b) To find the maximum length of MN, we need to determine the derivative of the
function that represents the length of MN, and set it equal to zero. This will give us
the x-value at which MN is a maximum.
𝑀𝑁 ′ = −2𝑥 + 10
For min/max 𝑓 ′ (𝑥) = 0:
0 = −2𝑥 + 10 Maximum value at 𝑓 ′ (𝑥) = 0
∴ 2𝑥 = 10
∴𝑥=5
Sub the 𝑥 value where the
∴ 𝑚𝑎𝑥 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑀𝑁 = −(5)2 + 10(5) − 4
maximum occurs into the
∴ 𝑚𝑎𝑥 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑀𝑁 = 21 𝑢𝑛𝑖𝑡𝑠 function that represents the
length of MN to find the
maximum length
2
Minima/Maxima problems
In minima/maxima problems you will be given a function or will have to formulate your own
function. You will have to, using calculus, determine the value of a variable that will optimise the
function. In order to determine the maximum or minimum you will have to determine the derivative
and set it equal to zero.
This is because any function that has a local minimum or maximum by virtue of the definition has
turning points, and as we recall the turning point is found at the point where 𝑓 ′ (𝑥) = 0.
Incredibly valuable in mastering this section is the realisation that any algebraic function can be
represented by a graph, this means that all our properties of graphs from the last three lessons are
still applicable here.
Steps to solving Minima/Maxima problems:
1. Draw a diagram if possible
2. If not given, determine an expression for the quantity that needs to be optimised, in terms
of one variable.
3. If there are 2 variables, eliminate one using the given information and simultaneous
equations
4. Write down 𝑓(𝑥) & 𝑓′(𝑥)
5. Set the derivative equal to 0 for minimum or maximum
6. Solve and answer question
1
, Example:
The functions 𝒇(𝒙) = 𝒙𝟐 − 𝟔𝒙 + 𝟓 and 𝒈(𝒙) = 𝟒𝒙 + 𝟏 are sketched below. 𝑴 is a point on 𝒈(𝒙)
and 𝑵 is a point on 𝒇(𝒙) such that 𝑵 lies directly underneath 𝑴, such that 𝑴𝑵 // to the 𝒚 −axis.
a) Find an expression for the length of 𝑴𝑵.
b) Determine the maximum length of 𝑴𝑵.
a) 𝑀𝑁 = 𝑔(𝑥) − 𝑓(𝑥) The length is the higher 𝑦-value less the lower 𝑦-value
∴ 𝑀𝑁 = 4𝑥 + 1 − (𝑥 2 − 6𝑥 + 5 )
∴ 𝑀𝑁 = 4𝑥 + 1 − 𝑥 2 + 6𝑥 − 5
∴ 𝑀𝑁 = −𝑥 2 + 10𝑥 − 4
b) To find the maximum length of MN, we need to determine the derivative of the
function that represents the length of MN, and set it equal to zero. This will give us
the x-value at which MN is a maximum.
𝑀𝑁 ′ = −2𝑥 + 10
For min/max 𝑓 ′ (𝑥) = 0:
0 = −2𝑥 + 10 Maximum value at 𝑓 ′ (𝑥) = 0
∴ 2𝑥 = 10
∴𝑥=5
Sub the 𝑥 value where the
∴ 𝑚𝑎𝑥 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑀𝑁 = −(5)2 + 10(5) − 4
maximum occurs into the
∴ 𝑚𝑎𝑥 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑀𝑁 = 21 𝑢𝑛𝑖𝑡𝑠 function that represents the
length of MN to find the
maximum length
2
