Nature of Points Nature of Points Applications Applications

Maxima, Minima, Point of Maxima, Minima, Point of InflectionInflection

RecapRecapWe saw how to find the coordinates of a turning point:We saw how to find the coordinates of a turning point:

DifferentiateDifferentiateSet f’(x) = 0Set f’(x) = 0Solve to find xSolve to find x

Substitute x into the original equation to find y.Substitute x into the original equation to find y.

f’(x) or f’(x) or dydy = 0 = 0 dxdxf’’(x) or f’’(x) or dd22yy Positive Positive

dxdx22

f’(x) or f’(x) or dydy = 0 = 0 dxdxf’’(x) or f’’(x) or dd22yy Negative Negative

dxdx22

Sketching curvesSketching curvesFind the stationary point(s):Find the stationary point(s): Find an expression for and put it equal to 0, then solve the resulting equation to find Find an expression for and put it equal to 0, then solve the resulting equation to find

the the xx co-ordinate(s) of the stationary point(s). co-ordinate(s) of the stationary point(s). Find and substitute each value of Find and substitute each value of xx to find the kind of stationary point(s). to find the kind of stationary point(s).

(+ suggests a minimum, – a maximum, 0 could be either or a point of inflection)(+ suggests a minimum, – a maximum, 0 could be either or a point of inflection) Use the curve’s equation to find the Use the curve’s equation to find the yy co-ordinate(s) of the stationary point(s). co-ordinate(s) of the stationary point(s).

Find the point(s) where the curve meets the axes:Find the point(s) where the curve meets the axes: Substitute Substitute xx = 0 in the curve’s equation to find the = 0 in the curve’s equation to find the yy co-ordinate of the point where the co-ordinate of the point where the

curve meets the curve meets the yy axis. axis. Substitute Substitute yy = 0 in the curve’s equation. If possible, solve the equation to find the = 0 in the curve’s equation. If possible, solve the equation to find the

xx co-ordinate(s) of the point(s) where the curve meets the co-ordinate(s) of the point(s) where the curve meets the xx axis. axis.

  

Sketch the curve, then use a graphic calculator to check.Sketch the curve, then use a graphic calculator to check.

In run mode

OPTN

F4 calc

F2 d/dx enter function

x value EXE will give

original g

,

radient

We can use the calculator to checkWe can use the calculator to check

2

2

2

F3 d/dx enter

function x value

EXE will gi e

,

vd y

dx

We can find the

y values using

graph mode

We can use the calculator to checkWe can use the calculator to check

In graph mode

Enter function F6 Draw

Shift F5 G solv F6

F1 Ycal

Enter x value EXE

Repeat for second x value

y

x

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3 22 3 12 15y x x x

Rules of How to Approach Applications of Rules of How to Approach Applications of Maxima and MinimaMaxima and Minima

1.If possible draw a diagram2.Use given information to form an equation about the situation.You may have to use other variables. One equation is about the given situation you are asked about. One relates y and x. Watch out for Pythagoras statements about perimeter, area, volume.3.Express y in terms of x (or x in terms of y whichever is simpler)

-I.e. Express one variable in terms of the other4.Differentiate and put equal to zero (you are finding the turning point of the graph either max or min)5.Solve to find x6.Re read the question7. Put x back in original equations to find other variables as required by the question8.Check that the value that you have found is in fact a

max or min as required. I.e. find

If it is <0, it is a max, If it is >0, it is a min

2

2( )

d yor f x

dx

A box with an open top can be made from a square sheet of A box with an open top can be made from a square sheet of cardboard with sides 12cm, by cutting a square of side x cardboard with sides 12cm, by cutting a square of side x

from each corner, and folding along the dotted lines.from each corner, and folding along the dotted lines.

What is the maximum volume?What is the maximum volume?

xx

12 cm

1

APPLICATION EXAMPLES WITH SOLUTIONS

Sides are 12cm :Sides are 12cm :

What is the length / width of the box in terms of x ?What is the length / width of the box in terms of x ?

L = 12 – 2xL = 12 – 2x

Volume of the box , Volume of the box , V = L * L * x V = L * L * x

= (12 – 2x)= (12 – 2x)22x x

= (144 – 24x – 24x + 4x= (144 – 24x – 24x + 4x22)x)x

= (144 – 48x + 4x= (144 – 48x + 4x22)x)x

=144x – 48x=144x – 48x22 + 4x + 4x33

2

Volume =144x – 48xVolume =144x – 48x22 + 4x + 4x33

Maximum volume - differentiate :Maximum volume - differentiate :dVdV = 144 – 96x + 12x = 144 – 96x + 12x22

dxdx

At maximum : At maximum : dVdV = 0 = 0dxdx

144 – 96x + 12x144 – 96x + 12x22 = 0 = 012(12 – 8x + x12(12 – 8x + x22) = 0) = 012(x – 6)(x – 2) = 012(x – 6)(x – 2) = 0So x = 6 or x = 2So x = 6 or x = 2

3

Find the nature of the turning Find the nature of the turning points points

Differentiate again :Differentiate again :

dVdV = 144 – 96x + 12x = 144 – 96x + 12x22

dxdxdd22VV = -96 +24x = -96 +24xdxdx22

Substitute in x = 6 Substitute in x = 6 dd22VV = -96 +24*6 = 48 = -96 +24*6 = 48dxdx22

> 0> 0 MinimumMinimumSubstitute in x = 2 Substitute in x = 2

dd22VV = -96 +24*2 = -48 = -96 +24*2 = -48dxdx22

< 0 < 0 MaximumMaximum

4

Maximum at x = 2Maximum at x = 2

Substitute in x = 2 to find maximum volume:Substitute in x = 2 to find maximum volume:

V =144x – 48xV =144x – 48x22 + 4x + 4x33

V = 144*2 – 48*2V = 144*2 – 48*222 + 4*2 + 4*233

V = 128 cmV = 128 cm33

NB don’t forget the unitsNB don’t forget the units

5

Typical ExampleTypical ExampleA farmer wishes to fence a paddock using an existing wall as one side of the paddock. He has 100m of fencing and wants to know the dimensions of the paddock enclosing the maximum area.

x

y

x

100 2y x 2100 2x x

2maximum area is 25 50 1250m 25x

100P 2P x y

2 100x y

A xy(100 2 )A x x

100 2y x

100 2(25)y

50y

Dimensions are 25m by 50m

If area is to be a maximum we must

differentiate and put it = 0

100 4dA

xdx

100 4 0x 4 100x

7

Example: Ex 18.3 # 6Example: Ex 18.3 # 6

60y x

y

y

x

If A is to be a maximum 0dA

dx

30y

2900cm

2

2

If we wish to prove this is a maximum, then we will

differentiate again 2 as this is <0 this a maximumd y

dx

#6) Find the maximum area of a rectangle

with a perimeter of 120cm

Perimeter 120cm2 2 120x y

60x y

A xy(60 )A x x

260A x x

60 2dA

xdx

60 2 0x 2 60x

30x

60y x

Question asked for the maximum

area A x y

A 30 30

x

6

y

x

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260A x x

X=side of paddock

Area

Of

paddock

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Drawing the graph for Ex 18.3 Question 6