A ball's height is $h(t) = -5t^2 + 20t + 1$ metres after $t$ seconds. Find its maximum height and when it reaches it.

A maximum height of $21$ m, after $2$ s.

A farmer has $60$ m of fencing for a rectangular pen against a barn wall, so only three sides need fence. What dimensions give the largest area?

$15$ m by $30$ m, with the $30$ m side opposite the wall, for an area of $450$ square metres.

A rock is thrown upward from the edge of a cliff. Its height above the water is $h(t) = -4.9t^2 + 14.7t + 49$ metres after $t$ seconds. Find the maximum height and when it happens, when the rock is back at cliff height, and when it hits the water.

A maximum of about $60.0$ m at $1.5$ s; back at cliff height at $3$ s; in the water at $5$ s.

Grade 11 · MCR3U · Functions

MCR3U: Quadratic Word Problems

Quadratic word problems come down to two questions. Either the problem wants a maximum or a minimum, which is the vertex, or it wants to know when a quantity reaches a value, which is an equation to solve. Decide which one it is before you write anything.

How It Works

Name the variable. Let $x$ stand for one quantity, with its units, and write the others in terms of it. A rectangle with perimeter $40$ m and width $x$ has length $20 - x$ .

Build the quadratic. Area is length times width, revenue is price times quantity sold, and the height of a thrown object is given as $h(t) = -4.9t^2 + v_0t + h_0$ in metres. Each time, a product of two linear expressions, or the physics, gives a quadratic.

Maximum or minimum: find the vertex. Complete the square, or use $x = -\frac{b}{2a}$ and substitute. The first coordinate says where the extreme value happens and the second says what it is.

Reaching a value: solve. Set the quadratic equal to that value, move everything to one side, then factor or use the quadratic formula. An object lands when its height is $0$ .

Reject what the context forbids. A quadratic usually gives two roots. Negative times, lengths or prices are not answers. A ball passes a given height twice, once going up and once coming down, so read which one the question wants. Say which root you keep and why.

Answer the question asked. Give the value in a sentence with units, rounded the way the question asks. Test questions often have several parts, such as the maximum height, when it happens and when the object lands, and each part is marked.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
The product of two consecutive positive integers is $132$ . Find them.
Show solution
1. Let the integers be $n$ and $n + 1$ : $n(n + 1) = 132$ , so $n^2 + n - 132 = 0$ .
2. Factor: $(n + 12)(n - 11) = 0$ , so $n = 11$ or $n = -12$ .
3. The integers are positive, so reject $-12$ .
Answer: $11$ and $12$ .
Easy
A ball's height is $h(t) = -5t^2 + 20t + 1$ metres after $t$ seconds. Find its maximum height and when it reaches it.
Show solution
1. Complete the square: $h = -5(t^2 - 4t + 4 - 4) + 1 = -5(t - 2)^2 + 21$ .
2. The vertex is $(2, 21)$ .
Answer: A maximum height of $21$ m, after $2$ s.
Medium
A farmer has $60$ m of fencing for a rectangular pen against a barn wall, so only three sides need fence. What dimensions give the largest area?
Show solution
1. Let $x$ be each side touching the wall. The side opposite the wall is $60 - 2x$ .
2. Area: $A = x(60 - 2x) = -2x^2 + 60x$ .
3. Complete the square: $A = -2(x^2 - 30x + 225 - 225) = -2(x - 15)^2 + 450$ .
4. The maximum is at $x = 15$ , with the long side $60 - 30 = 30$ m.
Answer: $15$ m by $30$ m, with the $30$ m side opposite the wall, for an area of $450$ square metres.
Medium
A rectangle's length is $3$ cm more than its width, and its area is $70$ square centimetres. Find its dimensions.
Show solution
1. Let the width be $x$ : $x(x + 3) = 70$ , so $x^2 + 3x - 70 = 0$ .
2. Factor: $(x + 10)(x - 7) = 0$ , so $x = 7$ or $x = -10$ .
3. A width cannot be negative, so $x = 7$ .
Answer: $7$ cm wide and $10$ cm long.
Test level
A rock is thrown upward from the edge of a cliff. Its height above the water is $h(t) = -4.9t^2 + 14.7t + 49$ metres after $t$ seconds. Find the maximum height and when it happens, when the rock is back at cliff height, and when it hits the water.
Show solution
1. Vertex: $t = -\frac{14.7}{2(-4.9)} = 1.5$ , and $h(1.5) = -4.9(2.25) + 22.05 + 49 = 60.025$ .
2. Cliff height is $h(0) = 49$ . Solve $-4.9t^2 + 14.7t + 49 = 49$ : $-4.9t(t - 3) = 0$ , so $t = 0$ or $t = 3$ .
3. The water is $h = 0$ . Divide $-4.9t^2 + 14.7t + 49 = 0$ by $-4.9$ : $t^2 - 3t - 10 = 0$ .
4. Factor: $(t - 5)(t + 2) = 0$ . Reject $t = -2$ , since the throw starts at $t = 0$ .
Answer: A maximum of about $60.0$ m at $1.5$ s; back at cliff height at $3$ s; in the water at $5$ s.

Learn It on Martian Lab

Martian Lab teaches this in the Problem Solving with Quadratics lesson of its MCR3U Quadratic Functions unit. Watch it, then take the Quadratic Functions practice tests to see if it stuck.

Watch the Problem Solving with Quadratics lesson free

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Free Quadratic Functions practice test, with solutions

More Quadratic Functions topics: Max/Min and Vertex Form · Linear-Quadratic Systems · Simplifying Radicals

More MCR3U topics: Inverse Functions · Transformations of Functions · Sine Law and the Ambiguous Case · Domain, Range and Function Notation · Simplifying Rational Expressions · Operations with Rational Expressions · Exponent Laws and Rational Exponents · Graphing Exponential Functions · Exponential Growth and Decay · Special Angles and the CAST Rule · Cosine Law and 3D Problems · Transformations of Sine and Cosine · Sinusoidal Word Problems · Arithmetic Sequences and Series · Geometric Sequences and Series · Pascal's Triangle and the Binomial Theorem · Simple and Compound Interest · Annuities: Future and Present Value

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