A. Characteristics of Functions · Quadratic Functions

Real-World Problems: Area, Projectiles and Numbers

A2.3 A2.2About 75 minutes

Learning goals

I can write a quadratic function in function notation from a description of an area, projectile or number problem.
I can find a maximum or minimum and say what it means in context.
I can use zeros to find when a quantity reaches a given value.
I can state a sensible domain and range for a real-world model.

Words to know

optimization
Finding the input that gives the greatest or least value of a function.

domain (in context)
The inputs that make sense for the situation, such as 0 ≤ t ≤ 4.10 for a ball in the air.

range (in context)
The outputs the situation actually produces.

projectile
An object that is thrown or launched and then moves under gravity alone.

The best rectangle, the highest throw

A farm family near Guelph has 60 m of fencing to build a rectangular pen against the long wall of a barn. How wide should the pen be to hold as many animals as possible? A baseball player throws a ball straight up. How high does it go, and for how long is it above 15 m? These are optimization and modelling problems. You will write each one as a quadratic function in function notation, decide what inputs make sense (the domain), and use the vertex and the zeros to answer.

A plan for word problems

Define the variable in words, with units ("let x be the width in metres").
Write the function, such as A(x) = …, h(t) = … or P(n) = ….
Restrict the domain to inputs that make sense (lengths and times cannot be negative).
Solve using the vertex (for a maximum or minimum) or the zeros (for when something equals a value).
Answer the question in a sentence with units, and check that it is reasonable.

Fencing three sides of a rectangle against a barn. The two sides of length x use 2x metres, leaving 60 − 2x metres for the side parallel to the barn.

Worked example

Example 1: Maximum area

A farm has 60 m of fencing to enclose a rectangular pen against a barn wall (no fence needed along the wall). Find the dimensions that give the maximum area.

Let x be the width (perpendicular to the barn) in metres. The side parallel to the barn is 60 − 2x.
A(x) = x(60 − 2x).
Domain: both sides must be positive, so x > 0 and 60 − 2x > 0, giving 0 < x < 30.
Zeros of A: x = 0 and x = 30. The axis is x = 15.
A(15) = 15(60 − 30) = 15 × 30 = 450.

Make the pen 15 m by 30 m (30 m along the barn), for a maximum area of 450 m².

Worked example

Example 2: A projectile, start to finish

A ball is thrown upward from a height of 2 m. Its height after t seconds is h(t) = −4.9t² + 19.6t + 2 metres. (a) Find the maximum height. (b) When does it land? (c) State the domain and range. (d) For how long is it above 15 m?

(a) t = −19.6/(2 × (−4.9)) = 2. h(2) = −19.6 + 39.2 + 2 = 21.6 m, after 2 s.
(b) Solve h(t) = 0: t = (−19.6 ± √(19.6² + 4(4.9)(2)))/(−9.8) = (−19.6 ± √423.36)/(−9.8) = (−19.6 ± 20.576)/(−9.8).
The positive solution is t = (−19.6 − 20.576)/(−9.8) ≈ 4.10 s. (The other solution, about −0.10, is before the throw, so we reject it.)
(c) Domain: 0 ≤ t ≤ 4.10 (from the throw to the landing). Range: 0 ≤ h ≤ 21.6.
(d) Solve h(t) = 15: −4.9t² + 19.6t − 13 = 0. D = 384.16 − 254.8 = 129.36, √D ≈ 11.374. t = (19.6 ∓ 11.374)/9.8 ≈ 0.84 or 3.16.
The ball is above 15 m between these times: 3.16 − 0.84 = 2.32 s.

Maximum 21.6 m at 2 s; lands after about 4.10 s; domain 0 ≤ t ≤ 4.10, range 0 ≤ h ≤ 21.6; above 15 m for about 2.32 s.

The height of the ball, h(t) = −4.9t² + 19.6t + 2, drawn only for 0 ≤ t ≤ 4.10. The shaded band shows the 2.32 s when the ball is above 15 m.

Why restrict the domain?

The equation h(t) = −4.9t² + 19.6t + 2 makes sense for every real t, but the situation does not: before t = 0 the ball had not been thrown, and after it lands it does not keep falling through the ground. A real-world model needs these physical restrictions on its domain and range.

Worked example

Example 3: Number problems

(a) Two numbers add to 24. What is their greatest possible product? (b) Two numbers differ by 10. What is their least possible product?

(a) Let the numbers be x and 24 − x. P(x) = x(24 − x), zeros 0 and 24, axis x = 12. P(12) = 12 × 12 = 144.
(b) Let the numbers be x and x + 10. P(x) = x(x + 10), zeros 0 and −10, axis x = −5. P(−5) = (−5)(5) = −25, with numbers −5 and 5.

(a) 144, from 12 and 12. (b) −25, from −5 and 5.

Worked example

Example 4: Using zeros, not the vertex

A 20 cm by 30 cm photo is placed in a frame with a border of equal width x all around. The total area of the photo and border is 1000 cm². How wide is the border?

The outside dimensions are 20 + 2x by 30 + 2x.
(20 + 2x)(30 + 2x) = 1000, so 600 + 100x + 4x² = 1000.
4x² + 100x − 400 = 0, or x² + 25x − 100 = 0.
x = (−25 ± √(625 + 400))/2 = (−25 ± √1025)/2. The positive solution is x ≈ (−25 + 32.02)/2 ≈ 3.51.

The border is about 3.5 cm wide.

Hands-on activity

String rectangles

You need: about 1.2 m of string or yarn, a ruler, tape and a flat table.

Tie the string into a loop that measures exactly 100 cm around (the extra string goes into the knot).
Use your fingers or tape to hold it in a rectangle 10 cm wide. Measure the length and compute the area.
Repeat for widths 15, 20, 25, 30 and 40 cm. Record width, length and area in a table.
Write the area as a function A(x) of the width x, and find its maximum algebraically. Does it match your table?
Bonus: untie the loop and mark off 100 cm of the string. Lay it against the edge of a book (like the barn wall) so only three sides use string. Which width gives the largest area now?

What to look for: the length is 50 − x, so A(x) = x(50 − x), with a maximum of 625 cm² for a 25 cm by 25 cm square. Against the book edge, A(x) = x(100 − 2x) is largest at x = 25 cm: a 25 cm by 50 cm rectangle with area 1250 cm².

Just for fun

A rectangle with a fixed perimeter always gets its biggest area by becoming a square. Squares have been bragging about it for centuries; rectangles just say squares are "too square".

Your turn

A rectangular pen against a wall uses 80 m of fence on three sides. Find the maximum area and the dimensions.

A rectangular garden is enclosed on all four sides by 100 m of fence. Find the maximum area.

A ball's height is h(t) = −5t² + 40t + 3 metres. Find the maximum height.

When does the ball from the last question land? Round to the nearest hundredth.

State a sensible domain and range for that ball's height function.

Two numbers add to 30. Find the greatest possible product.

Two numbers differ by 8. Find the least possible product and the numbers.

More practice: warm-up

A rectangle has perimeter 40 m and width x. Write its area as a function of x.

What is the maximum area of a rectangle with perimeter 40 m?

h(t) = −5(t − 3)² + 50. What is the maximum height and when?

Why can't the domain of a thrown-ball model include negative values of t?

Write the product of two numbers that add to 10 as a function of one of them, x.

More practice: on your own

120 m of fence makes a rectangular enclosure divided into two equal pens by one fence parallel to the width. Find the maximum total area.

For h(t) = −4.9t² + 19.6t + 2 (Example 2), when is the ball above 10 m?

A rectangle's length is 3 cm more than twice its width, and its area is 65 cm². Find its dimensions.

A diver leaves a 10 m platform. Her height above the water is h(t) = −4.9t² + 3t + 10 metres. When does she enter the water?

Find the diver's maximum height above the water.

Challenge

A rectangle has one corner at the origin and the opposite corner on the line y = 12 − 2x in the first quadrant. Find the maximum area.

A farmer has 600 m of fence to enclose a rectangular field and split it into three equal pens with two fences parallel to the width. Find the maximum total area.

A ball is thrown upward from 1.5 m with initial speed v m/s, so h(t) = −4.9t² + vt + 1.5. What value of v makes the maximum height exactly 20 m?

Check your understanding

Quiz

A pen against a wall uses 40 m of fence on three sides, A(x) = x(40 − 2x). What is the maximum area?

A ball's height is h(t) = −5t² + 20t, and it lands at t = 4. What is a sensible domain?

Two numbers add to 18. Their greatest possible product is:

h(t) = −5t² + 30t. What is the maximum height?

In A(x) = x(60 − 2x) for a pen, why must x be less than 30?

What is the range of h(t) = −4.9t² + 19.6t + 2 for the ball in Example 2?

For parents and teachers

Big idea: Expectation A2.3: solve real-world problems with quadratic functions represented in function notation. It also uses domain and range (A1.3 in the sibling unit Functions and Equivalent Expressions) in context.

How to help: Ask your teen to draw a labelled sketch for every problem and to say in words what each variable means.

Common difficulties:

Using 2x + 2L for fencing when one side is a wall (or a dividing fence is added).
Giving the width when the question asks for the area, or the time when it asks for the height.
Keeping a negative root for a time or length.
Forgetting that −4.9t² in physics comes from gravity (half of 9.8 m/s²), with heights in metres and times in seconds.