Optimisation questions involve finding the best solution to a problem. Usually, that is the maximum or minimum value of a function. Questions on this topic require you to take information from a practical problem and write this as a function, then use differentiation to find the maximum or minimum value of this function by solving where the gradient is equal to zero. Often questions on this topic seem to have two variables, but you need to take a piece of information that fixes one of the variables and substitute this to create a function with one variable (although HL students might have to use Implicit Differentiation!)
On this page, you should learn about
- solving optimisation problems using differentiation
The following videos will help you understand all the concepts from this page
Optimum Can
In the following video we look at an example of using differentiation to find the best possible can for a drinks manufacturer.
A drinks manufacturer wants a design for a new can for their mini drinks collection, volume=170ml.Your job is to find the can that will contain a volume of 170ml = 170cm3 ≈54πcm3 and that will minimize the amount of aluminium used.
It might help you to visualize the can by playing with the following applet
Notes from the video
The following is an example of an optimisation problem
A rectangle ABCD is drawn so that its lower vertices are on the x-axis and its upper vertices are on the curve \(f(x)=\sqrt{6-3x^2}\) as shown in the following diagram
Let OA = x
The area of this rectangle is denoted by A.
(a) Write down an expression for A in terms of x.
(b) Find the maximum value of A.
You may find the following applet useful to get an understanding of the problem.
Notes from the video
Here is a quiz that practises setting up the problem before you use Calculus to solve. You are strongly recommended to have something to write with whilst doing these questions. You may want to draw sketches of shapes and graphs to help you think through the problems. You are not required to do any Calculus in this quiz!
START QUIZ!
120m of fencing is used to make an enclosure against a wall where x represents the width of the enclosure.
What is the fomula for the area? A of the pen
A rectangle has area 20cm². If one side of the rectangle is x, find a formula for the perimeter of the rectangles
Perimeter = \(x + \frac{20}{x}+x + \frac{20}{x}\)
P = \(2x + \frac{40}{x}\)
An open box is made from a piece of paper 20cm by 20cm by removing a square from each corner and folding.
If x represents the side of the square, what is the formula for the volume, V of the box?
Length = 20 - 2x
Width = 20 - 2x
Height = x
Volume = length x width x height
Volume = (20 - 2x)(20 - 2x)x
Volume = x(20 - 2x)²
An open box is made from a piece of paper 30cm by 20cm by removing a square from each corner and folding.
If x represents the side of the square, what is the formula for the volume, V of the box?
Length = 30 - 2x
Width = 20 - 2x
Height = x
Volume = length x width x height
Volume = (30 - 2x)(20 - 2x)x
A piece of wire 50cm long is bent into the shape of a rectangle. If x represents the length of the side of the rectangle, what is the formula for the area, A of the rectangle created?
Width = 25-x
Area = x(25 - x)
A closed cuboid box has has volume 1000cm3
If the base is square with side x cm, find the total surface area of the cuboid
Volume = x²h
1000 = x²h
\(h=\frac{1000}{x^2}\)
Total surface area \(=2\times x^2+4 \times x\times \frac{1000}{x^2}\)
Total surface area \(=2x^2+\frac{4000}{x}\)
ABC is a right-angled triangle. The sum of the lengths AB and AC is 10cm.
Find the area of the triangle if AB = x
We can use Pythagoras' Theorem to find the missing length
Area =\(\frac{x}{2}\sqrt{100-20x}\)
A rectangle is drawn so that its lower vertices lie on the x axis and its upper vertices lie on the curve
y = a² - x²
Find an expression for the area of the rectangle.
It is a good idea to draw a sketch of the graph.
The width of the rectangle is 2x
The height of the rectangle is the value of the function a² - x²
Therefore, the area = 2x(a² -x²)
A rectangle is drawn so that its lower vertices lie on the x axis and its upper vertices lie on the curve
y = f(x)
\(f(x)=\left\{\begin{matrix}e^x\ \ \ \ \ \ \ \ x<0\\e^{-x}\ \ \ \ \ \ \ \ x\geq0\\\end{matrix}\right.\)
Find an expression for the area of the rectangle.
It is a good idea to draw a sketch of the graph.
The width of the rectangle is 2x
The height of the rectangle is the value of the function \(e^{-x}\)
Therefore, the area = \(2xe^{-x}\)
A half cylinder has volume = 400
Find a formula for the A, surface area of the solid in terms of r, the radius of the corss section
Let the length of the cylinder = l
Volume = \(\frac{\pi r^2}{2}l\)
\(\frac{\pi r^2}{2}l=400\)
\(l=\frac{800}{\pi r^2}\)
The surface area of the solid is made up of
2 semi-circular ends = \(2\times\frac{\pi r^2}{2}\)
+
rectangular base = \(2rl\)
+
curved suface = \(\pi rl\)
Total surface area = \(\pi r^2+2rl+\pi rl\)
\(A=\pi r^2+2r\frac{800}{\pi r^2}+\pi r\frac{800}{\pi r^2}\)
\(A=\pi r^2+\frac{1600}{\pi r}+\frac{800}{ r}\)
A container is made from a cylinder and a hemisphere. The radius of the cylinder is r m and the height is h. The volume of the container is \(45\pi\)
a) Find an expression for the height of the cylinder in terms of r
b) Show that the surface area of the container, \(A=\frac{5 \pi r^2}{3}+\frac{90 \pi}{r}\)
c) Hence, find the values of r and h that give the minimum surface area of the container
* Volume of a sphere = \(\frac{4}{3}\pi r^3\)
** Surface area of a sphere = \(4\pi r^2\)
Hint
Full Solution
The diagram below shows the graph of the functions f(x) = sinx and g(x) = 2sinx
A rectangle ABCD is placed in between the two functions as shown so that B and C lie on g , BC is parallel to the x axis and the local minima of the function f lies on AD.
Let NA = x
a) Find an expression for the height of the rectangle AB
b) Show that the area of the rectangle, A can be given by A = 4xcosx - 2x
c) Find \(\frac{dA}{dx}\)
d) Find the maximum value of the area of the rectangle.
Hint
Full Solution
How much of Optimisation have you understood?
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