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.

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 Millers 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