Everything Maths Grade 12

CHAPTER 7. DIFFERENTIAL CALCULUS 7.6

P�(a) = 10− 2 a

0 = 10− 2 a

2 a = 10

a =

10

2

a = 5

Substitute into (7.26) tosolve for the width.

b = 10−a

= 10− 5

= 5

Step 4 : Write the final answer

The product is maximised when a and b are both equal to 5.

Example 14: Optimisation Problems

QUESTION

Michael wants to starta vegetable garden, which he decides to fenceoff in the shape of a

rectangle from the rest of the garden. Michael has only 160 m of fencing, so he decides to use

a wall as one border ofthe vegetable garden. Calculate the width andlength of the garden

that corresponds to largest possible area that Michael can fence off.

wall

garden

length,

l

width, w

SOLUTION

Step 1 : Examine the problem and formulate the equations that are required

The important pieces ofinformation given are related to the area and modified

perimeter of the garden.We know that the area of the garden is:

A = w.l (7.24)

We are also told that thefence covers only 3 sides and the three sides should add

up to 160 m. This can be written as:

160 m = w + l + l (7.25)