Algebra 1 · A.SSE.B.3B · A.CED.A.1

Maximum, minimum and projectile problems

A situation has a best choice, the most area, the least cost, the highest point of a throw. How do you find it exactly, rather than by trying numbers, and how do you tell when the question wants the best value and when it wants something else?

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Using the quadratic formulaAlgebra 1 Vertex form: reading, converting and graphingAlgebra 1 The quadratic formulaGrades 9 to 10

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Fitting a quadratic to dataAlgebra 2

Where you are 12 in Quadratics

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Quadratics12

▴ 7 above The quadratic formula4 Vertex form: reading, converting and graphing Factored form

Maximum, minimum and projectile problems

Fitting a quadratic to data

When you reach for it

Some questions ask for the best of something. What price brings in the most money? How should 240 meters of fence be laid out to hold the most ground? How high does a thrown ball go? Each has one best answer, and trying numbers only ever gets close to it.

When the quantity you want to make large or small is a quadratic in one variable, its graph is a parabola, and a parabola turns exactly once. That turning point, the vertex, is the answer. Its first coordinate says when or where the best happens: the time, the width, the price. Its second coordinate says how much: the height, the area, the money.

Reach for this move whenever a word problem says greatest, least, most, fewest, maximum, minimum, largest, smallest or highest, and the quantity involved is, or can be written as, a quadratic. The same models also carry questions that are not about a best value at all: when does the ball land, which price brings in exactly this much. Those are answered by a root, not the vertex, and the first part of the procedure is telling the two apart.

The most common source of such a quadratic is a thrown object, whose height is a quadratic in time. Projectile motion has its own part of the procedure below.

The procedure

Quadratics in context: a root or the vertex?

Some quantities rise and then fall, or fall and then rise: the height of a thrown ball, the profit as a price goes up, the area of a pen made from a fixed length of fence. Each is modeled by a quadratic. The work is knowing which piece of the parabola answers the question asked.

The vertex answers “how much at most” or “how little at least”. It also answers “when” or “where” that best value happens, through its first coordinate. Greatest height, most profit, largest area, least cost: all vertex questions.

A root answers “when does it reach” or “what makes it equal”. When does the ball land? Which prices bring in exactly 2275 dollars? Which width gives an area of 2400 square meters? Each sets the quadratic equal to a number and solves, and the answers are the roots of what is left. A problem that starts from a known area and asks for the lengths, an area and dimension problem, is of this kind, usually solved by factoring; any root that makes a length negative is thrown out.

Both at once. A projectile problem often asks for the peak, a vertex question, and the landing, a root question, about the same ball.

Building the model runs the other way. Fitting a quadratic to data starts from measurements: three points with different x values, not all on one line, fix a parabola, so you find a, b and c from the data and only then read its vertex or roots. Finding them means solving a system in three unknowns.

Finding a greatest or least value

Take the fence problem from the opening. A rectangular field is to be fenced with 240 meters of fencing, and one more fence runs across the middle, parallel to two of the sides, to make two pens. Which dimensions give the most ground?

Name the variable and the quantity to make best. The quantity here is the area A. It depends on the two side lengths, so call them x and y, with x the length of the three fences that run the same way (the two ends and the divider).

Figure 1Three pieces of length x and two of length y use all 240 meters.

Use the other facts to get down to one variable. A vertex belongs to a graph of one variable, so every fact in the problem that ties the lengths together is used to remove one. All the fence together is 3x + 2y = 240, so \displaystyle y = 120 - \frac{3}{2}x. Then

\begin{aligned} A &= xy \\ &= x\left(120 - \tfrac{3}{2}x\right) \\ &= -\tfrac{3}{2}x^2 + 120x \end{aligned}

Read the sign of a to know which kind of point the vertex is. Here \displaystyle a = -\frac{3}{2} < 0, so the parabola opens downward and the vertex is its highest point, the greatest area. If a were positive, the vertex would be a least value instead.

Find the first coordinate from the axis of symmetry. The vertex lies on the axis of symmetry, \displaystyle x = -\frac{b}{2a}:

x = -\frac{120}{2\left(-\frac{3}{2}\right)} = \frac{120}{3} = 40

Find the second coordinate by substituting. The best value is what the quantity equals there:

A = -\tfrac{3}{2}(40)^2 + 120(40) = 2400

Check that the vertex is allowed. Both lengths must be positive, so x > 0 and \displaystyle 120 - \frac{3}{2}x > 0, that is 0 < x < 80. The vertex x = 40 lies inside, so it is a field that can be built. A vertex outside the allowed values is not the answer, as Example 4 shows.

Answer the question that was asked, in its units. The question asked for dimensions, so find y as well: y = 120 - 60 = 60. The fences across are 40 meters, the long sides 60 meters, and the field holds 2400 square meters.

The vertex can come other ways too. Vertex form gives it by reading, and when the quadratic comes as a product, as areas and revenues often do, the vertex sits halfway between the two roots. Here \displaystyle A = x\left(120 - \frac{3}{2}x\right) is zero at x = 0 and at x = 80, and halfway is 40.

Projectile motion

Throw a ball straight up and three questions come with it. How high does it get? When does it land? When is it at some particular height, say level with a window? Each has a number for an answer, and all three come from one formula. Near the ground, and ignoring air, the height of the ball t seconds after it leaves your hand is

h = -16t^2 + v_0 t + h_0

in feet, where v_0 is the speed it was launched upward in feet per second and h_0 is the height it was launched from. In meters the same law reads h = -4.9t^2 + v_0 t + h_0.

That is a quadratic in t, so its graph is a parabola opening downward, and each question is a question about the parabola. The highest point is the vertex. The landing is a root. A given height is where a level line cuts the curve. Reach for this whenever something is thrown, launched, kicked or dropped and only its height matters.

Take a ball thrown straight up at 64 feet per second from a hand 6 feet above the ground.

Write the model, and know what each term is doing. Here v_0 = 64 and h_0 = 6:

h = -16t^2 + 64t + 6

Each piece has a job. At t = 0 the first two terms vanish and h = 6, so h_0 is where the ball starts. With no gravity the ball would rise at a steady 64 feet every second, and its height would be the straight line h = 64t + 6. Gravity pulls it below that line by 16t^2 feet: 16 feet after one second, 64 after two, four times as far in twice the time. That is where the minus sign and the square come from.

Figure 2Without gravity the ball would follow the straight line. Gravity pulls it 16t^2 feet below that line, so the gap is 16 after one second and 64 after two.

The 16 is half of 32, the pull of gravity in feet per second per second. In meters it is half of 9.8, which is why the metric model has 4.9.

For the highest point, find the vertex. A parabola is symmetric about its axis, and the axis of h = at^2 + bt + c is at \displaystyle t = -\frac{b}{2a}. Here

t = -\frac{64}{2(-16)} = 2

and substituting gives the height there:

h = -16(4) + 64(2) + 6 = 70

Because a = -16 is negative the parabola opens downward, so the vertex is the highest point, not the lowest. The ball peaks 70 feet up, 2 seconds after release. In vertex form the model is h = -16(t - 2)^2 + 70, which says the same thing at a glance: the squared term can only take height away, and it takes none at t = 2.

For the landing, solve h = 0 and keep the positive root. The ground is height 0. From the vertex form,

\begin{aligned} -16(t - 2)^2 + 70 &= 0 \\ (t - 2)^2 &= \frac{70}{16} \\ t &= 2 \pm \frac{\sqrt{70}}{4} \end{aligned}

The minus sign gives t \approx -0.09, a time before the ball was thrown, when the model does not describe anything. The plus sign gives the landing, t \approx 4.09 seconds.

Figure 3The vertex is the peak. The positive root is the landing.

Any of the methods in choosing how to solve a quadratic can solve h = 0. Factor if the numbers allow it, take square roots when the model is in vertex form, and use the quadratic formula otherwise.

For a given height H, solve h = H. Move H across and solve the quadratic that is left. Two positive roots mean the ball passes that height twice, going up and coming down. A repeated root means H is exactly the peak. No real roots mean the ball never gets that high, which the discriminant can tell you before you solve anything. And if H is below the launch height, one root is negative: the ball passes H only once, on the way down.

Check against the picture. Every answer must lie between release and landing. Any two times at the same height sit symmetrically about the peak time, so their average is the peak time.

Worked examples

Example 1

A least cost from a given model

A workshop finds that its cost, in dollars, of making x tables in a day is C(x) = 2x^2 - 48x + 400. How many tables should it make each day to keep the cost as low as possible, and what is that cost?

Solution

The model is already a quadratic in one variable, with a = 2, b = -48 and c = 400. Since a > 0 the graph opens upward, so the vertex is the lowest point and gives the least cost.

x = -\frac{-48}{2(2)} = 12

C(12) = 2(144) - 48(12) + 400 = 112

Making 12 tables a day costs 112 dollars, the least possible.

Check. One table either side should cost the same, and more: C(11) = 242 - 528 + 400 = 114 and C(13) = 338 - 624 + 400 = 114.

Example 2

The price that brings in the most

A school play sells 300 tickets at 8 dollars each. For every 1 dollar the price goes up, it expects to sell 25 fewer tickets. What price brings in the most money, and how much is that?

Solution

Let n be the number of 1 dollar rises. The price is then 8 + n dollars and the number sold is 300 - 25n. Money in, the revenue, is price times number sold:

\begin{aligned} R &= (8 + n)(300 - 25n) \\ &= -25n^2 + 100n + 2400 \end{aligned}

Choosing the variable is the real step here: the price and the number sold both change, but both are fixed once n is.

Since a = -25 < 0 the vertex is the highest point:

n = -\frac{100}{2(-25)} = 2

Two rises give a price of 10 dollars and sales of 300 - 50 = 250 tickets, so the revenue is 10 \cdot 250 = 2500 dollars, against 2400 at the old price.

Check. The factored form is zero when n = -8 (a free ticket) and when n = 12 (nobody buys), and halfway between is \displaystyle \frac{-8 + 12}{2} = 2.

Example 3

A least value after expanding

Two numbers add to 20. Make the sum of their squares as small as possible.

Solution

Call one number x; the other is then 20 - x. The quantity to make small is

S = x^2 + (20 - x)^2

This is a quadratic, but its a and b cannot be read until it is expanded:

\begin{aligned} S &= x^2 + 400 - 40x + x^2 \\ &= 2x^2 - 40x + 400 \end{aligned}

Now a = 2 > 0, so the vertex is the least value:

x = -\frac{-40}{2(2)} = 10

The numbers are 10 and 10, and the least sum of squares is 100 + 100 = 200.

Check. Moving one apart, 9 and 11 give 81 + 121 = 202, which is more.

Example 4

When the vertex is not allowed

A pen is made against a wall that is only 30 meters long, using 100 meters of fence for the other three sides. The side opposite the wall must fit along it. What is the largest area?

Solution

Call each side meeting the wall x. The side opposite the wall is 100 - 2x, and the area is

A = x(100 - 2x) = -2x^2 + 100x

The vertex is at \displaystyle x = -\frac{100}{2(-2)} = 25, with A = 1250. But at x = 25 the opposite side would be 100 - 50 = 50 meters, longer than the wall. The vertex is not an allowed pen.

The allowed widths come from the conditions. The opposite side must fit: 100 - 2x \le 30, so x \ge 35. It must also be positive, so x < 50.

Figure 4The top of the arch is not an allowed pen. Over the allowed strip the area only falls, so the best allowed pen is at the strip’s left edge.

The whole allowed range lies to the right of the vertex, and there the downward parabola only falls. So the largest allowed area is at the left end, x = 35:

A = 35(100 - 70) = 35 \cdot 30 = 1050

The pen is 35 meters out from the wall and 30 meters along it, holding 1050 square meters. The 1250 at the vertex would need a longer wall.

Example 5

When a root answers instead

The school play of Example 2 has revenue R = -25n^2 + 100n + 2400 dollars after n rises of 1 dollar. (a) Which ticket prices bring in exactly 2275 dollars? (b) Can any price bring in 2600 dollars?

Solution

Neither part asks for a best value. Part (a) asks what makes the revenue equal a given number, so it is answered by solving an equation, and its answers are roots.

(a) Set R = 2275 and move everything to one side:

\begin{aligned} -25n^2 + 100n + 2400 &= 2275 \\ -25n^2 + 100n + 125 &= 0 \\ n^2 - 4n - 5 &= 0 \end{aligned}

Dividing by -25 changes no solution. Then (n - 5)(n + 1) = 0, so n = 5 or n = -1. A negative number of rises is a price cut of 1 dollar, which is a price the play could charge, so both roots are kept, provided the rule of 25 tickets a dollar holds for a cut as well as a rise. The prices are 8 + 5 = 13 dollars and 8 - 1 = 7 dollars.

Check. At 13 dollars, 300 - 125 = 175 tickets sell, and 13 \cdot 175 = 2275. At 7 dollars, 300 + 25 = 325 sell, and 7 \cdot 325 = 2275. The two roots average to \displaystyle \frac{5 + (-1)}{2} = 2, the vertex of Example 2: two prices giving the same revenue sit symmetrically about the best one.

(b) Here the vertex does decide the answer. The most the play can bring in is the 2500 dollars found at the vertex in Example 2, so no price brings in 2600. The equation agrees: -25n^2 + 100n - 200 = 0 is n^2 - 4n + 8 = 0, with discriminant 16 - 32 = -16, which is negative, so there is no real n.

Example 6

Up to the top and back to the hand

A ball is tossed straight up at 40 feet per second from a height of 5 feet. When is it highest, how high does it go, and when is it back at the height it was thrown from?

Solution

The launch speed is v_0 = 40 and the launch height is h_0 = 5, so

h = -16t^2 + 40t + 5

The highest point is the vertex, since the parabola opens downward. Its time is

t = -\frac{40}{2(-16)} = \frac{40}{32} = 1.25

and its height is found by substituting that time:

\begin{aligned} h &= -16(1.25)^2 + 40(1.25) + 5 \\ &= -25 + 50 + 5 = 30 \end{aligned}

The ball peaks at 30 feet, 1.25 seconds after it is thrown. As a check, it climbed 25 feet, and Example 11 shows the climb is always \displaystyle \frac{v_0^2}{64}, here \displaystyle \frac{1600}{64} = 25.

It is back at 5 feet when -16t^2 + 40t + 5 = 5. The 5s cancel, leaving -16t^2 + 40t = -8t(2t - 5) = 0, so t = 0, the throw itself, or t = 2.5. That is exactly twice the peak time. The two times at height 5 sit symmetrically about t = 1.25, so the ball takes as long to fall back to the hand as it took to climb from it.

Example 7

Launched from the ground

A model rocket is launched straight up from the ground at 96 feet per second, so its height is h = -16t^2 + 96t. When does it land, and how high does it get?

Solution

It lands when h = 0. There is no constant term, so t is a common factor:

-16t^2 + 96t = -16t(t - 6) = 0

A product is zero when one of its factors is, so t = 0 or t = 6. The first is the launch, when the rocket is also on the ground. The second is the landing: 6 seconds.

The peak is halfway between two times at the same height, and the two roots are two times at height 0. So it comes at t = 3, and

h = -16(9) + 96(3) = -144 + 288 = 144

The rocket lands after 6 seconds and reaches 144 feet. Launched from the ground, it spends exactly as long going up as coming down.

Example 8

A landing time that needs a square root

A ball is thrown up at 32 feet per second from a height of 10 feet. When does it hit the ground?

Solution

The model is h = -16t^2 + 32t + 10, and the landing is where h = 0. The equation does not factor over the integers, so write it in vertex form first. The vertex is at \displaystyle t = \frac{32}{32} = 1, where h = -16 + 32 + 10 = 26, so

h = -16(t - 1)^2 + 26

Now set it to zero and undo the square:

\begin{aligned} 16(t - 1)^2 &= 26 \\ (t - 1)^2 &= \frac{26}{16} \\ t &= 1 \pm \frac{\sqrt{26}}{4} \end{aligned}

The minus sign gives about -0.27, before the throw, so it is rejected. The ball lands at \displaystyle t = 1 + \frac{\sqrt{26}}{4} \approx 2.27 seconds.

The quadratic formula agrees. With a = -16, b = 32, c = 10, the discriminant is 32^2 - 4(-16)(10) = 1664 = 64 \cdot 26, and

t = \frac{-32 \pm 8\sqrt{26}}{-32} = 1 \mp \frac{\sqrt{26}}{4}

Example 9

When is it at a given height?

An arrow is shot straight up at 80 feet per second from 5 feet above the ground, so h = -16t^2 + 80t + 5. (a) When is it 69 feet up, and for how long is it above that height? (b) Does it ever reach 110 feet?

Solution

(a) Set h = 69 and move everything to one side:

\begin{aligned} -16t^2 + 80t + 5 &= 69 \\ -16t^2 + 80t - 64 &= 0 \\ t^2 - 5t + 4 &= 0 \end{aligned}

Dividing by -16 changes no solution and leaves small numbers. Then (t - 1)(t - 4) = 0, so t = 1 or t = 4. The arrow passes 69 feet on the way up after 1 second and on the way down after 4. Between those times it is higher, so it spends 4 - 1 = 3 seconds above 69 feet.

Figure 5The arrow passes 69 feet twice, once going up and once coming down. It never reaches 110 feet.

(b) The two times average to 2.5, which is therefore the peak time, and the peak height is

h = -16(6.25) + 80(2.5) + 5 = 105

The arrow never gets above 105 feet, so it never reaches 110. The equation says so too: -16t^2 + 80t - 105 = 0 has discriminant 80^2 - 4(-16)(-105) = 6400 - 6720 = -320, which is negative, so there is no real time at which h = 110.

Example 10

Reading the launch speed from the flight

In meters, rounding 4.9 to 5 to keep the arithmetic whole, a ball thrown up from a balcony 6 meters high has height h = -5t^2 + v_0 t + 6. It is 21 meters up both 1 second and 3 seconds after release. Find its launch speed, its greatest height and when it lands.

Solution

The two times at the same height lie symmetrically about the peak, so the peak is at their average, t = 2. The vertex of -5t^2 + v_0 t + 6 is at \displaystyle t = \frac{v_0}{10}, so

\frac{v_0}{10} = 2 \qquad\Longrightarrow\qquad v_0 = 20

The ball was launched at 20 meters per second. Check: h(1) = -5 + 20 + 6 = 21 and h(3) = -45 + 60 + 6 = 21. The greatest height is

h(2) = -20 + 40 + 6 = 26

meters, so in vertex form h = -5(t - 2)^2 + 26. It lands when this is 0:

\begin{aligned} (t - 2)^2 &= \frac{26}{5} \\ t &= 2 + \sqrt{\frac{26}{5}} \approx 4.28 \end{aligned}

taking the positive root, since the other is about -0.28. The ball lands after about 4.28 seconds.

Example 11

How fast must it be thrown?

A ball is launched straight up from the ground. How fast must it go to reach exactly 64 feet? And to reach a height four times as great?

Solution

Work with a general launch speed v_0 and a general launch height h_0. The peak time is \displaystyle t = \frac{v_0}{32}, and substituting it into h = -16t^2 + v_0 t + h_0 gives the peak height:

\begin{aligned} h &= -16\left(\frac{v_0}{32}\right)^2 + \frac{v_0^2}{32} + h_0 \\ &= -\frac{v_0^2}{64} + \frac{2v_0^2}{64} + h_0 \\ &= h_0 + \frac{v_0^2}{64} \end{aligned}

So a ball always climbs \displaystyle \frac{v_0^2}{64} feet above its launch point. From the ground h_0 = 0, and reaching 64 feet needs

\frac{v_0^2}{64} = 64 \qquad\Longrightarrow\qquad v_0^2 = 4096

so v_0 = 64 feet per second, taking the positive root because the launch is upward.

The climb grows with the square of the speed. Four times the height, 256 feet, needs v_0^2 = 64 \cdot 256 = 16384, so v_0 = 128: only twice as fast.

Boundaries

The quantity must be a quadratic in one variable. A problem with two unknowns needs a fact linking them, used to remove one, before there is a parabola to have a vertex. If no such fact exists, the method has nothing to work on.

The sign of a says which question can be answered. A parabola that opens upward has a least value and no greatest one; its outputs grow without end. So “the most” of an upward quadratic has no answer over all inputs, and “the least” of a downward one has none either.

The vertex is the answer only when it is allowed. Lengths must be positive, counts must be whole, a price rise may be capped, a wall may be short. If the vertex lies outside the allowed inputs, the best allowed value is at the end of the allowed range nearest the vertex, as in Example 4. If the variable must be a whole number and the vertex is not one, the best whole number is one of the two either side of it, whichever gives the better value.

A root must be allowed too. A negative time, a negative length or a price below zero is mathematics, not the situation, and is thrown away. A negative number of price rises, on the other hand, is a price cut and may be perfectly real, as in Example 5. Whether a root is kept is a question about the situation, not about the sign.

The answer is only as good as the model. “25 fewer tickets for every dollar” is a guess about buyers. The vertex is exact for the model, not a promise about the world.

For a projectile: no air. The model assumes gravity is the only force. For a heavy ball over a few seconds that is close. For a feather, a shuttlecock or a long-range shell, air resistance changes the answer, and the true curve is no longer a parabola.

For a projectile: only the height. The horizontal axis is time, not distance, so the graph is not the path of a ball thrown straight up, which goes up and down along one vertical line. For a ball thrown at an angle the height still follows this model, with v_0 the upward part of the launch speed, and the sideways motion is handled on its own by projectile motion in parametric form.

For a projectile: only between launch and landing. The formula describes the ball for 0 \le t \le t_{\text{land}}. The negative root and the parts of the parabola below the ground are mathematics, not the ball, and are thrown away.

Units must match. 16 goes with feet and seconds, 4.9 with meters and seconds. Put a speed in meters per second into the feet model and every answer is wrong.

Ground means height zero. h is measured up from some level, usually the ground. If the ball lands on a roof 12 feet up, the landing is h = 12, not h = 0.

Where this goes next

The habit this lesson builds, asking what a feature of the graph means in the situation, carries into modeling with polynomials and every function you model after this. In fitting a quadratic to data the parabola is built from measurements first, measured heights of a ball or measured costs, and then its vertex and roots are read in exactly this way.

The fixed-perimeter rectangle is an old friend: the square encloses the most, and a quadratic now proves it (Exercise 36).

The -16t^2 was taken on trust here. In calculus it is derived: free fall starts from a constant acceleration and integrates twice, and the v_0 t and h_0 appear as the two constants of integration. And once the sideways motion is added, parametric equations follow a ball thrown at an angle across the field as well as up and down.

Most quantities worth making best are not quadratics: a box folded from card has a volume cubic in the fold. For those, optimization in calculus finds the turning points through critical points, where the graph is momentarily flat, as a parabola is at its vertex.

Exercises

Read the answer from a model

The daily cost, in dollars, of making x units is C(x) = 2x^2 - 80x + 1000. How many units make the cost least, and what is it?
A shop’s weekly profit, in dollars, when it charges p dollars for an item is P(p) = -10p^2 + 600p - 5000. What price gives the greatest profit, and what is it?
A cable hangs between two towers. Its height above the road, x meters from the first tower, is y = 0.01x^2 - 2x + 110 meters. How close does it come to the road, and where?

Build the model, then find the best value

Two numbers differ by 8. Find the smallest possible product, and the numbers that give it.
A rectangle is fenced on all four sides with 200 meters of fence. Find its largest area and its dimensions.
A farmer fences three sides of a rectangular plot along a river with 120 meters of fence; the river is the fourth side. Find the largest area and the dimensions that give it.
A rectangle is fenced with 300 meters of fence and divided into three pens by two more fences parallel to its ends. Find the largest total area.
A theater sells 400 tickets at 20 dollars each. Each 1 dollar rise in price loses 10 sales. What price brings in the most money, and how much is it?
Two numbers add to 16. Find the smallest possible sum of their squares.
A metal sheet 30 cm wide is bent up by x cm along each long edge to make a gutter with a rectangular cross-section. What value of x gives the largest cross-section, and what is its area?
An orchard has 50 trees to the acre, each giving 600 apples. Each extra tree planted per acre lowers the yield of every tree by 10 apples. How many trees per acre give the most apples, and how many apples is that?

When the vertex is not allowed

The plot in Exercise 6 lies on a river bank where only 40 meters of bank can be used, so the side along the river is at most 40 meters. Find the largest area now.
A workshop’s weekly profit, in dollars, from making x chairs is P(x) = -x^2 + 60x - 500, but it can make at most 20 chairs a week. Find the greatest profit it can make.
A supplier’s cost, in dollars, of an order of x crates is C(x) = x^2 - 40x + 500, and every order is at least 25 crates. Find the least cost of an order.

A root or the vertex

The theater of Exercise 8 has revenue R = (20 + n)(400 - 10n) dollars after n rises of 1 dollar. Which ticket prices bring in exactly 8750 dollars?
The rectangle of Exercise 5 is fenced with 200 meters of fence. Which dimensions give it an area of exactly 2400 square meters? Can it enclose 2600 square meters?
The shop of Exercise 2 has profit P(p) = -10p^2 + 600p - 5000. At which prices does it break even, making a profit of exactly 0?
For each question, say whether the vertex or a root of the model answers it, without solving. (a) When does a thrown ball land? (b) What is the greatest profit? (c) At what price is the profit zero? (d) How high does the ball go? (e) For how long is the ball above 20 feet?

Thrown objects: reading the model

A ball’s height in feet is h = -16t^2 + 32t + 5. From what height was it launched, at what upward speed, and how high is it after 2 seconds?
Write the model for a ball thrown straight up at 40 feet per second from 4 feet above the ground, and find its height after 1 second.

Thrown objects: the highest point

h = -16t^2 + 72t + 4. When is the ball highest, and how high is it then?
In meters, h = -5t^2 + 30t + 2. When is the ball highest, how high is it then, and when is it back at the 2 meters it started from?
h = -16(t - 1.5)^2 + 40. Find the greatest height, when it happens, and the launch height.
A ball is thrown straight up from the ground at 56 feet per second. How high does it go?

Thrown objects: landing

Find when each object hits the ground. Give exact answers, and decimals to two places where the answer is not whole.

h = -16t^2 + 64t
h = -16t^2 + 16t + 32
h = -16t^2 + 24t + 10
A stone is dropped, not thrown, from the top of a cliff 144 feet high.
In meters, h = -5t^2 + 10t + 40.
h = -16(t - 2)^2 + 80

Thrown objects: a given height

h = -16t^2 + 96t. When is the object 128 feet up, and for how long is it higher than that?
h = -16t^2 + 48t + 4. When is the ball 20 feet up?
h = -16t^2 + 64t + 4. Does the ball ever reach 70 feet?
How fast must a ball be launched from the ground to reach exactly 144 feet?
A ball thrown straight up from 6 feet is 18 feet up at t = 0.5 and again at t = 1.5. When is it highest, what was its launch speed, and how high does it go?

Reasoning

Show that of all rectangles with perimeter p, the one with the largest area is a square, and find that area.
Show that two numbers with a fixed sum s have their largest product when they are equal, and that the product is then \displaystyle \frac{s^2}{4}.
A quadratic model has a < 0, and its vertex lies to the right of every allowed input. Explain why its greatest allowed value is at the right-hand end of the allowed range.
Explain why the question “what is the greatest cost?” has no answer for the model C(x) = 2x^2 - 48x + 400 of Example 1, if x may be any number.
Show that a ball launched from the ground at speed v_0 takes as long to come down as it took to go up.
Explain why doubling the launch speed of a ball thrown from the ground makes it go four times as high.
The model for a ball thrown from 5 feet up has a root near t = -0.15. What does that root mean, and why is it thrown away?
A friend graphs h = -16t^2 + 48t and says the ball flew in an arch across the park. What is wrong with reading the graph that way?

Answers

20 units, at a cost of 200 dollars. \displaystyle x = -\frac{-80}{2(2)} = 20, and 800 - 1600 + 1000 = 200.
A price of 30 dollars, for a profit of 4000 dollars. \displaystyle p = -\frac{600}{2(-10)} = 30, and -9000 + 18000 - 5000 = 4000.
It comes within 10 meters of the road, at x = 100. \displaystyle x = -\frac{-2}{2(0.01)} = 100, and 100 - 200 + 110 = 10. Here a > 0, so this is a least height.
The smallest product is -16, from -4 and 4. The product is x(x + 8) = x^2 + 8x, which opens upward with vertex at x = -4, and -4 \cdot 4 = -16.
The largest area is 2500 square meters, a 50 by 50 square. The sides are x and 100 - x, so A = x(100 - x), with roots 0 and 100 and vertex halfway, at x = 50.
The largest area is 1800 square meters: 30 meters out from the river and 60 meters along it. A = x(120 - 2x) has roots 0 and 60, so the vertex is at x = 30.
The largest area is 2812.5 square meters. Four fences of length x and two long sides of length y give 4x + 2y = 300, so y = 150 - 2x and A = x(150 - 2x). The roots are 0 and 75, so x = 37.5, y = 75, and A = 37.5 \cdot 75.
A price of 30 dollars, bringing in 9000 dollars. With n rises the revenue is R = (20 + n)(400 - 10n), which expands to -10n^2 + 200n + 8000, so n = 10: 300 tickets at 30 dollars.
The smallest sum is 128, from 8 and 8. S = x^2 + (16 - x)^2 = 2x^2 - 32x + 256, with vertex at x = 8.
x = 7.5 cm, for an area of 112.5 square centimeters. The base is 30 - 2x, so A = x(30 - 2x), with roots 0 and 15 and vertex at x = 7.5; the base is then 15 cm.
55 trees per acre, for 30250 apples. With n extra trees the total is (50 + n)(600 - 10n), which expands to -10n^2 + 100n + 30000, so n = 5, and 55 \cdot 550 = 30250.
The largest area is now 1600 square meters, a 40 by 40 plot. The side along the river is 120 - 2x \le 40, so x \ge 40. The vertex x = 30 is not allowed, and the area falls to the right of it, so the best allowed plot has x = 40.
The greatest profit is 300 dollars, at x = 20. The vertex is at x = 30, outside the allowed range, and the profit rises all the way up to x = 20: -400 + 1200 - 500 = 300.
The least cost is 125 dollars, for 25 crates. The vertex is at x = 20, below the smallest allowed order, and the cost rises from there, so the smallest allowed order is cheapest: 625 - 1000 + 500 = 125.
Prices of 25 dollars and 35 dollars. -10n^2 + 200n + 8000 = 8750 becomes n^2 - 20n + 75 = 0, which is (n - 5)(n - 15) = 0. Check: 25 \cdot 350 = 8750 and 35 \cdot 250 = 8750. The two roots average to 10, the vertex of Exercise 8.
40 meters by 60 meters. x(100 - x) = 2400 becomes x^2 - 100x + 2400 = 0, which is (x - 40)(x - 60) = 0, and either root gives the same rectangle. It cannot enclose 2600: the most it can hold is 2500, at the vertex.
At 10 dollars and at 50 dollars. -10p^2 + 600p - 5000 = 0 divides to p^2 - 60p + 500 = 0, which is (p - 10)(p - 50) = 0. Between those prices the shop makes a profit; outside them it loses money.
(a) A root, of h = 0. (b) The vertex, its second coordinate. (c) A root, of P = 0. (d) The vertex, its second coordinate. (e) Two roots, of h = 20; the time above is the gap between them. If the ball starts above 20 feet, one root is negative and the time above runs from release to the positive root.
Launched from 5 feet at 32 feet per second; after 2 seconds it is 5 feet up. h_0 is the constant term and v_0 the coefficient of t, and h(2) = -64 + 64 + 5 = 5.
h = -16t^2 + 40t + 4, and after 1 second the ball is 28 feet up: -16 + 40 + 4 = 28.
Highest at t = 2.25, at 85 feet. \displaystyle t = \frac{72}{32} = 2.25, and h = -81 + 162 + 4 = 85.
Highest at t = 3, at 47 meters; back at 2 meters at t = 6. \displaystyle t = \frac{30}{10} = 3 and h = -45 + 90 + 2 = 47; then -5t^2 + 30t = -5t(t - 6) = 0 gives t = 6, twice the peak time.
The greatest height is 40 feet, at t = 1.5; the launch height is 4 feet. The vertex is (1.5, 40), and h(0) = -16(2.25) + 40 = 4.
49 feet, at t = 1.75. The peak is \displaystyle \frac{v_0^2}{64} = \frac{3136}{64} = 49 above the ground.
t = 4 seconds. -16t(t - 4) = 0, and t = 0 is the launch.
t = 2 seconds. -16(t^2 - t - 2) = -16(t - 2)(t + 1), and t = -1 is before the throw.
\displaystyle t = \frac{3 + \sqrt{19}}{4} \approx 1.84 seconds. Divide by -2: 8t^2 - 12t - 5 = 0, and the formula gives \displaystyle \frac{12 \pm \sqrt{304}}{16} = \frac{3 \pm \sqrt{19}}{4}; the minus sign is negative.
t = 3 seconds. A dropped stone has v_0 = 0, so h = -16t^2 + 144, and 16t^2 = 144 gives t^2 = 9.
t = 4 seconds. -5(t^2 - 2t - 8) = -5(t - 4)(t + 2).
\displaystyle t = 2 + \sqrt{5} \approx 4.24 seconds. (t - 2)^2 = 5, and \displaystyle 2 - \sqrt{5} is negative.
At t = 2 and t = 4, so it is above 128 feet for 2 seconds. -16t^2 + 96t - 128 = 0 divides to t^2 - 6t + 8 = 0, which is (t - 2)(t - 4) = 0.
At \displaystyle t = \frac{3 - \sqrt{5}}{2} \approx 0.38 going up and \displaystyle t = \frac{3 + \sqrt{5}}{2} \approx 2.62 coming down. -16t^2 + 48t - 16 = 0 divides to t^2 - 3t + 1 = 0.
No. The peak is at t = 2, where h = -64 + 128 + 4 = 68, so the ball never gets above 68 feet.
96 feet per second. \displaystyle \frac{v_0^2}{64} = 144 gives v_0^2 = 9216.
Highest at t = 1, launched at 32 feet per second, reaching 22 feet. The peak is midway between 0.5 and 1.5; then \displaystyle \frac{v_0}{32} = 1, and h(1) = -16 + 32 + 6 = 22.
If one side is x, the other is \displaystyle \frac{p}{2} - x, and the area is \displaystyle A = x\left(\frac{p}{2} - x\right) = -x^2 + \frac{p}{2}x. This opens downward with vertex at \displaystyle x = \frac{p}{4}, where the other side is \displaystyle \frac{p}{2} - \frac{p}{4} = \frac{p}{4} as well, so the rectangle is a square, with area \displaystyle \frac{p^2}{16}.
With numbers x and s - x, the product x(s - x) = -x^2 + sx opens downward, so its vertex, at \displaystyle x = \frac{s}{2}, is the largest. Then the other number is also \displaystyle \frac{s}{2}, and the product is \displaystyle \frac{s}{2} \cdot \frac{s}{2} = \frac{s^2}{4}.
A downward parabola rises all the way up to its vertex and falls after it. If the vertex is to the right of every allowed input, the whole allowed range is on the rising side, so each input gives more than any input to its left. The largest input allowed therefore gives the largest value. If that end were itself excluded, there would be no greatest value at all.
The coefficient of x^2 is 2 > 0, so the parabola opens upward. Its vertex is its lowest point, and on either side the cost climbs without limit, since 2x^2 outgrows -48x for large x. Whatever cost is named, some x gives more, so there is no greatest cost.
From the ground h = -16t^2 + v_0 t = -t(16t - v_0), which is zero at t = 0 and \displaystyle t = \frac{v_0}{16}. The peak is at \displaystyle t = \frac{v_0}{32}, exactly halfway, so the rise and the fall each take \displaystyle \frac{v_0}{32} seconds.
From the ground the peak height is \displaystyle \frac{v_0^2}{64}. Replacing v_0 by 2v_0 gives \displaystyle \frac{4v_0^2}{64}, four times as much, because the speed is squared.
It is where the parabola, extended backwards past the throw, would meet the ground. The model only describes the ball from release, t = 0, onward, so a negative time is not part of the flight.
The horizontal axis is time, not distance across the park. The graph says how high the ball is at each moment. A ball thrown straight up goes up and comes down along one vertical line, and the arch is only the record of its height rising and falling over 3 seconds.