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UNIT 3 : QUADRATIC FUNCTIONS & EQUATIONS
LESSON
3: PROBLEM SOLVING WITH QUADRATIC
FUNCTIONS
Finding Maximum or Minimum Values of
Quadratic Functions:
To find
maximum or minimum values of a quadratic function, we must put it in vertex
form: y = a(x h)2 + k . This is done by completing
the square
or by
using the x=-b/2a method.
Quadratic
Function Problems (Max/Min Problems)
Number
Problems:
Example 2:
Find two
numbers which differ by 24 such that their product is a minimum.
Solution:
ΰ Determine
what is unknown or what you are asked to find?
Assign variables to the unknowns. Here we are asked
to find two numbers.
Let the first number be x and the second
number be y.
ΰ
Determine the quantity to be maximized or minimized.
We are asked to find the minimum product of the
two numbers. Hence our formula is:
ΰ Write this formula as a function of one variable.
We are told the numbers differ by
24. This yields the secondary
relation x y = 24
Isolating x we obtain x = y + 24
Area
Problems:
Example 3:
Farmer Al
has 240 ft. of fencing available. He
wishes to enclose a rectangular garden with this fence. One side borders a stream bank and requires
no fence.
Find the
dimensions he should use to enclose a field of maximum area.
Solution:
ΰ Determine
what is unknown or what you are asked to find?
Assign variables to the unknowns. Here the
dimensions of the field are unknown.
Let the length be x and the width be y.
Stream bank
y y
x
ΰ
Determine the quantity to be maximized or minimized.
We are asked to maximize the area. Hence our formula is:
ΰ Write this formula as a function of
one variable. Since the available fencing is240 ft,
we have a secondary relation between the two variables.
Economics
Problems:
There
are two important formulas at work here.
First Revenue or Income from sale of
goods will be the product of Quantity Sold and Price.
The
second formula is : Profit is found by
subtracting costs (expenses) from revenue or income.
Example 4:
Last
month Millers sold power walking shoes at $150 per pair. At this price they sold 60 pairs of
shoes. A marketing research survey
shows
that for
each $10 price increase, sales will drop by 3 pairs per month.
a) Determine the selling price that will
maximize revenue and state the maximum revenue at this price.
b) If the wholesale price of the shoes to
Miller is $80.00, find the selling price that will maximize profits.
Solution:
ΰ The key step in these problems is the Let statement:
a) Let the number
of price increases
($10.00) be x.
Revenue (last month) = Quantity x
Price = 60 x 150
New Price:
There will be x increases of $10.00 each. Hence the new price will be (150 + 10x)
New quantity
Each $10 price increase produces a sales drop of 3 pairs. Hence new quantity sold is (60 3x).
New Revenue = Quantity x Price = (60 3x)(150
+ 10x)
b) To maximize profits, we use the second
equation above, namely Profit = Revenue Cost
.
Since we already have a function for revenue, we must modify it by
subtracting costs.
Since the wholesale price/pair is $80.00,
the cost function C(x) will be:
.
The
solution proceeds by forming the profit function as follows.
b) Find maximum height by putting in vertex
form ie complete the square
d)
Example 6:
A ball is thrown in the air from ground level. Its height in metres at any time t
seconds is given by h(t) = -4.9t2
+ 24t + 0.4.
a) Find the initial height of the ball.
b) Find the maximum height attained by the ball and when it occurs.
c) When will the ball hit the ground.
d) Draw the graph showing time on the x-axis.
Solution:
a) Initial height means t = 0.
b) Find maximum height by putting in vertex
form ie complete the square
c)
When
the ball hits the ground, height is 0.
d)
Height
Time
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