# Maximum & Minimum Applications

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Maximum & Minimum Applications Using differentiation techniques to determine maximum values, optimal solutions, of minimum values. Max & Min applications

Skills: Extract relevant information from a word problem, form an equation, differentiate and solve the problem. Class Notes | Blank Notes Class Notes | Blank Notes Graphs: Ex15.06

#### NuLake IAS p93 to 97 Cutter cross section

## Remember the steps:

### Form an equation Differentiate Make the derivative = 0 to find the max & min Solve to find 'x' Answer the question

###### Example:
 Find the minimum surface area of a cuboids (box) which has a square end, and a volume of 800cm3 Steps: 1) Find an equation for surface area A = 2x2 + 4xL We only want one variable so eliminate L using the volume equation 2) Volume equation - rearrange to be L = V = Lx2 800 = Lx2 L = 800/x2 3) Substitute L = 800/x2 back into the Area Equation A = 2x2 + 4xL A = 2x2 + 4x800/x2 A = 2x2 + 3200/x 4) Differentiate the Area equation A = 2x2 + 3200x-1 A' = 4x - 3200x-2 5) Make the derivative = 0 and solve for x (multiply by x2) 0 = 4x - 3200x-2 0 = 4x3 - 3200 3200 = 4x3 800 = x3 x = 3√800 = 9.28cm 6) Calculate the surface area A = 2x2 + 3200/x A = 9.28 2 + 3200/9.28 A = 517.1cm3

#### back to top ###### Example:
Find the minimum surface area of a cylinder which has a volume of 330cm3

What are the dimensions of the cylinder? # Steps:

1) Find an equation for surface area

2 circles + rectangle
A = 2пr2 + 2пrL

We only want one variable so eliminate L using the volume equation

2) Volume equation - rearrange to be L =

V = пr2L
330 = пr2L
L = 330/пr2

3) Substitute L = 330/2пr2 back into the Area Equation

A = 2пr2 + 2пrL
A = 2пr2 + 2пr330/пr2
A = 2пr2 + 660/r

4) Differentiate the Area equation

A = 2пr2 + 660r-1
A' = 4 пr - 660r-2

5) Make the derivative = 0 and solve for r
(multiply by r 2)

0 = 4 пr - 660r-2
660r-2 = 4пr
165r-2 = пr
52.52 = r3
r = 3√52.52 = 3.745cm (1dp)

6) Calculate the surface area

A = 2п3.7452 + 660/3.745
A = 264.4cm3 (1dp)

Dimensions

r = 5.5cm (1dp)

Length

L = 330/пr2
L = 330/п3.7452
L = 7.7cm