Quadratic Optimization

Intuition

A great many ordinary questions ask for the best: the largest garden you can fence, or the price that brings in the most money. Others ask for the highest a ball rises, or the cheapest way to run a machine. Last lesson you learned that the graph of $y = ax^2 + bx + c$ is a parabola with a single turning point, the vertex. You also learned that this vertex is the lowest point of the curve when it opens up and the highest point when it opens down. That is exactly what an optimization question is after. If the quantity you care about is a quadratic, its best value is not something you hunt for by testing numbers. That best value sits at the vertex, and you already know how to find the vertex. The one habit to build from the start is to keep two different things apart: the input that produces the best result, and the best result itself. They are almost never the same number, and mixing them up is the single most common way these problems go wrong.

Learning goals

Locate the best value at the vertex $x = -\tfrac{b}{2a}$
Read the sign of $a$ to tell a maximum from a minimum
Separate where the best occurs from how good it is
Model a problem as a quadratic, then answer what was asked
Apply it to fixed perimeter, revenue and projectile height

Why the vertex is the best value

Take any quadratic quantity written as $y = ax^2 + bx + c$ . From the parabola lesson you can find its vertex in two moves: the input is

x = -\frac{b}{2a},

and the value there is whatever you get by substituting that $x$ back into the equation. If the quadratic is already in vertex form $y = a(x - h)^2 + k$ , it is even quicker, because the vertex is $(h, k)$ . The value $k$ at the vertex is then sitting right in front of you.

Which of the two, a maximum or a minimum, you have found is decided entirely by the sign of $a$ . A positive $a$ opens the parabola upward into a valley, so the vertex is the lowest point and $y$ has a minimum there. A negative $a$ opens it downward into a hill, so the vertex is the highest point and $y$ has a maximum. There is no separate rule to memorize for word problems. Instead, read the sign of the leading coefficient, and you know whether the vertex is a floor or a ceiling.

Why the vertex holds the largest or smallest value#

Write the quadratic in vertex form by completing the square, $y = a(x - h)^2 + k$ , where the vertex is $(h, k)$ . Everything hangs on the squared piece $(x - h)^2$ . A square is never negative, so $(x - h)^2 \ge 0$ , and it equals $0$ at exactly one input, $x = h$ .

Suppose $a > 0$ . Multiplying a nonnegative number by a positive number keeps it nonnegative, so $a(x - h)^2 \ge 0$ , and therefore

y = a(x - h)^2 + k \ge k.

The value $y$ can never dip below $k$ , and it actually reaches $k$ only when the square is $0$ , that is at $x = h$ . So $k$ is the smallest value the quadratic ever takes: a minimum, occurring at the vertex.

Now suppose $a < 0$ . Multiplying a nonnegative number by a negative number makes it nonpositive, so $a(x - h)^2 \le 0$ , and therefore $y \le k$ . This time $y$ can never rise above $k$ , and it reaches $k$ only at $x = h$ , so $k$ is the largest value: a maximum, again at the vertex.

Either way the extreme value is $k$ , the height of the vertex, and it is reached at the single input $x = h = -\tfrac{b}{2a}$ . The sign of $a$ alone decides which extreme it is. This is why optimizing a quadratic is never a search through many inputs. There is one special point, and the two-step vertex calculation lands on it directly.

This picture is worth holding in your mind for the rest of the lesson. The quantity you want to make as large or as small as possible traces a parabola, and you are simply locating its peak or its valley.

Optimizing a quadratic quantity means locating the vertex of its parabola. Here the quantity opens downward, so the vertex is the peak, its greatest value. Read the answer off two axes: the input that achieves the best result is directly below the peak, and the best value itself is directly across from it. They are different numbers.

Before any word problems, practice the pure mechanics on a bare quadratic, keeping the input and the value in separate boxes in your head.

Worked example 1 The maximum value of $y = -2x^2 + 12x - 7$

The leading coefficient is $a = -2$ . Since $a < 0$ the parabola opens downward, so its vertex is the highest point and the quantity has a maximum. Find the input where that happens with $x = -\tfrac{b}{2a}$ , reading $a = -2$ and $b = 12$ :

x = -\frac{b}{2a} = -\frac{12}{2(-2)} = -\frac{12}{-4} = 3.

The maximum occurs at the input $x = 3$ . To get the maximum value, substitute $x = 3$ back into the equation:

y = -2(3)^2 + 12(3) - 7 = -18 + 36 - 7 = 11.

So the maximum value of $y$ is $11$ , and it happens at $x = 3$ . Notice how different those two numbers are. If a question asks for the maximum value, the answer is $11$ ; the number $3$ is only where it occurs. Answering $3$ to “what is the maximum” is the classic slip this lesson is built to prevent.

Check your understanding

Find the minimum value of $y = x^2 - 6x + 1$ .

-8

3

8

1

A method for optimization problems

Real problems arrive as sentences, not as $y = ax^2 + bx + c$ . Turning the words into a quadratic reuses the same modeling discipline you built for quadratic applications. There is one change at the end: instead of setting the quantity equal to a target and solving, you find its vertex.

Name the variable. Pick a letter for the quantity you get to choose (a width, a price, a time), and write every other quantity in the problem in terms of it.
Build the quadratic. Express the thing you want to make biggest or smallest (an area $A$ , a revenue $R$ , a height $h$ , a product $P$ ) as a quadratic in your variable.
Find the vertex. Compute the optimizing input with $x = -\tfrac{b}{2a}$ , and read the sign of $a$ to confirm you have a maximum ( $a < 0$ ) or a minimum ( $a > 0$ ).
Answer the exact question, with units. Decide whether the problem wants the optimizing input, the optimal value, or both, then substitute back as needed and state the result in a sentence. Check that the answer is physically sensible (a length is positive, a count is a whole number).

Step 4 is where the input-versus-value distinction earns its keep. “What price maximizes revenue” wants the input; “what is the greatest revenue” wants the value. A question that says “find the dimensions of the largest pen” wants the inputs, and the area they produce is a separate follow-up. Read the final sentence of the problem carefully and give back precisely what it asks.

Greatest area for a fixed amount of fence

Area problems are the heart of this topic, because area is a product of two lengths. When a fixed perimeter ties those two lengths together, the product of the two becomes a quadratic. A fixed supply of fence is the usual setup.

Worked example 2 Largest rectangular pen from 40 metres of fence

A farmer has $40$ metres of fencing to enclose a rectangular pen. What dimensions enclose the greatest area, and what is that area?

Name the width $x$ (in metres). The four sides are two widths and two lengths, and they use all $40$ metres, so $2x + 2(\text{length}) = 40$ , which gives a length of $20 - x$ . Area is length times width:

A = x(20 - x) = 20x - x^2.

This is a quadratic in $x$ with $a = -1$ and $b = 20$ . Since $a < 0$ the parabola opens downward, so its vertex is a maximum. The optimizing width is

x = -\frac{b}{2a} = -\frac{20}{2(-1)} = 10.

The width is $10$ metres, and the length is $20 - x = 10$ metres, so the best pen is a $10$ by $10$ square. Its area is the value at the vertex:

A = 10(20 - 10) = 10 \cdot 10 = 100.

The greatest area is $100$ square metres. For a fixed perimeter, the maximum-area rectangle always turns out to be a square, which the algebra has just confirmed.

That last claim is worth catching in the act rather than taking on trust, because it rests on many different rectangles sharing a single perimeter. The figure below reports both quantities for whatever rectangle you build, so you can pin one of them and watch the other.

Hold the perimeter at $20$ by keeping the width and the height adding to $10$ . Now walk along that family: $1$ by $9$ , then $2$ by $8$ , $3$ by $7$ , $4$ by $6$ , and finally $5$ by $5$ . The perimeter readout never leaves $20$ , while the area climbs $9$ , $16$ , $21$ , $24$ , $25$ . The square is the peak. Keep going past it to $6$ by $4$ and $7$ by $3$ . The areas then run back down through $24$ and $21$ , mirroring the values on the way up. That mirror is the axis of symmetry of $A = x(10 - x)$ , and the top of the climb is its vertex, which $x = -\tfrac{b}{2a}$ locates without any of this walking. One more setting is worth visiting for contrast: build $9$ by $1$ , then $3$ by $3$ . The first reports perimeter $20$ and area $9$ , the second perimeter $12$ and area $9$ . Equal areas, and nothing like equal perimeters. So one perimeter can go with many areas, and one area with several perimeters. That is exactly why an optimization problem has to say which of the two it is holding fixed before “the best rectangle” means anything at all.

Rectangle explorer

Width Height

A rectangle 3 units wide and 7 units tall. Perimeter 20 units. Area 21 square units.

A rectangle sized in unit squares, reporting its perimeter and its area as two separate readings. Rectangles can share a perimeter while differing in area, and share an area while differing in perimeter, so a family that holds one of the two fixed can be walked through and compared on the other.

Change the setup so the pen backs onto a wall, and only three sides need fencing. The wall gives one side for free, so the fence is split differently, but the method does not budge.

Worked example 3 Largest pen against a wall with 60 metres of fence

A rectangular garden is built against a long straight wall, so the wall forms one side and fencing is needed only for the other three. With $60$ metres of fence, what is the largest area?

Let $x$ be the length of each side that runs out from the wall (there are two of these). The remaining fence forms the side parallel to the wall, so that side is $60 - 2x$ . The area is

A = x(60 - 2x) = 60x - 2x^2.

Here $a = -2$ and $b = 60$ , and $a < 0$ signals a maximum. The optimizing input is

x = -\frac{b}{2a} = -\frac{60}{2(-2)} = -\frac{60}{-4} = 15.

Each side out from the wall is $15$ metres, and the side parallel to the wall is $60 - 2(15) = 30$ metres. The greatest area is

A = 15(30) = 450.

So the largest garden is $450$ square metres, from a $15$ by $30$ rectangle. Notice the shape is no longer a square: freeing up one side with the wall changes the balance. In this setup the side parallel to the wall comes out twice as long as each of the other two.

The three-sided pen of Worked Example 3. The wall supplies the fourth side, so the 60 metres of fence covers two sides of length x running out from the wall and one side of length 60 minus 2x parallel to it. The area A = x(60 minus 2x) is a downward parabola, largest at the vertex x = 15, which gives a 15 by 30 garden.

The same product structure runs through pure number problems. Two numbers with a fixed sum have a product that is a quadratic. With the sum held fixed, that product is largest when the two numbers are equal, which is the number version of the square being the best rectangle.

Check your understanding

Two positive numbers add up to $16$ . What is the greatest possible value of their product?

60

32

64

8

Greatest revenue

Money problems create a quadratic whenever raising the price drives down the number sold. Revenue is price times quantity, and if the quantity falls off in a straight line as the price climbs, that product is again a quadratic with a peak.

Worked example 4 The ticket price that maximizes revenue

A theater sells $200$ tickets per show at $10$ dollars each. A survey predicts that every $1$ dollar increase in the ticket price will sell $10$ fewer tickets. What price brings in the greatest revenue?

Let $x$ be the number of $1$ dollar increases. Then the price is $10 + x$ dollars, and the number of tickets sold is $200 - 10x$ . Revenue is price times quantity:

R = (10 + x)(200 - 10x).

Expand so you can see the coefficients:

R = 2000 - 100x + 200x - 10x^2 = -10x^2 + 100x + 2000.

Here $a = -10$ and $b = 100$ , and $a < 0$ means the vertex is a maximum. The optimizing number of increases is

x = -\frac{b}{2a} = -\frac{100}{2(-10)} = -\frac{100}{-20} = 5.

Five increases means the best price is $10 + 5 = 15$ dollars, at which the theater sells $200 - 10(5) = 150$ tickets. The greatest revenue is the value at the vertex:

R = (15)(150) = 2250.

So a ticket price of $15$ dollars maximizes revenue, bringing in $2250$ dollars. The question asked for the price, so $15$ dollars is the headline answer; the revenue of $2250$ dollars is a separate figure. The raw vertex input $x = 5$ (the number of increases) is neither of those, just a step along the way.

A projectile’s highest point

The falling-object model from the applications chapter is a quadratic in time, so its highest point is a vertex. Near the surface of the Earth an object thrown into the air has height

h = -16t^2 + v_0\,t + h_0 \qquad (\text{height in feet, time } t \text{ in seconds}),

with $h_0$ the starting height and $v_0$ the initial upward speed. In metric units the same model reads $h = -4.9t^2 + v_0\,t + h_0$ with the height in metres. The leading coefficient is negative in both, so the graph of height against time is a downward parabola, and its vertex is the top of the flight. The input at the vertex is the time the object peaks; the value there is the maximum height. Keep those two apart exactly as before.

Worked example 5 Maximum height of a launched ball

A ball is launched upward from a $6$ foot platform with an initial speed of $80$ feet per second. So its height in feet after $t$ seconds is $h = -16t^2 + 80t + 6$ . Find the maximum height and the time it occurs.

The coefficients are $a = -16$ , $b = 80$ , and $c = 6$ . Because $a < 0$ , the vertex is the highest point. The time at the peak is

t = -\frac{b}{2a} = -\frac{80}{2(-16)} = -\frac{80}{-32} = 2.5.

The ball peaks at $t = 2.5$ seconds. That is the time, not the height. Substitute $t = 2.5$ back to get the maximum height:

\begin{aligned} h &= -16(2.5)^2 + 80(2.5) + 6 \\ &= -16(6.25) + 200 + 6 = -100 + 200 + 6 = 106. \end{aligned}

The maximum height is $106$ feet, reached at $t = 2.5$ seconds. A question that asks “how high” wants $106$ feet; a question that asks “when” wants $2.5$ seconds. The model hands you both, one as the vertex input and one as the vertex value.

Check your understanding

A stone thrown upward has height $h = -16t^2 + 64t + 5$ feet after $t$ seconds. At what time does it reach its maximum height?

69

feet B

2

seconds C

4

seconds D

64

feet

Least cost and other minimums

When the leading coefficient is positive the parabola opens upward, and the vertex becomes a lowest point rather than a highest one. Everything about the method is identical; only the sign of $a$ has flipped, and with it the word “maximum” becomes “minimum”. Costs that fall and then rise are a natural home for these.

Worked example 6 The production level that minimizes cost

A workshop finds that its daily cost in dollars to make $x$ units of a product is $C = 2x^2 - 40x + 500$ . How many units should it make to minimize the cost, and what is that minimum cost?

The coefficient $a = 2$ is positive, so the parabola opens upward and the vertex is the lowest point, a minimum. The optimizing number of units is

x = -\frac{b}{2a} = -\frac{-40}{2(2)} = \frac{40}{4} = 10.

Making $10$ units minimizes the cost. The minimum cost is the value there:

C = 2(10)^2 - 40(10) + 500 = 200 - 400 + 500 = 300.

So the workshop should make $10$ units, and the least possible daily cost is $300$ dollars. The positive leading coefficient is the whole reason this is a minimum. The cost drops as production rises toward $10$ units, bottoms out there, and climbs again beyond it.

Everything in this lesson comes back to one image. Whatever the quantity, area, revenue, height, product, or cost, if it is a quadratic then its graph is a parabola. The answer you want is the peak or the valley of that parabola. Find the vertex, check the sign of $a$ to know which one it is, and then read off precisely the input or the value the question asks for.

Common mistakes

Reporting the input when the question wants the value (or the reverse). The vertex has two coordinates. The input $x = -\tfrac{b}{2a}$ tells you where the best happens; the value found by substituting back tells you how good it is. Decide which the sentence asks for before you answer.
Stopping at $x = -\tfrac{b}{2a}$ . That is only the first coordinate of the vertex. For a maximum height, a greatest area, or a least cost, you still have to substitute back to get the value.
Guessing maximum versus minimum. Do not assume every problem is a maximum. Read the sign of $a$ : $a < 0$ opens down to a maximum, $a > 0$ opens up to a minimum. A cost or a sum being minimized has $a > 0$ .
Fencing a wall problem like a four-sided one. When a wall supplies one side, the fixed fence covers only three sides, so the relationship is $2x + (\text{opposite side}) = \text{fence}$ , not the full perimeter. Writing all four sides double-counts the free wall.
Building the wrong product. Revenue is price times quantity, and area is one length times the other. Multiply the two quantities the problem actually ties together, and write the second in terms of the first before expanding.
Dropping units or the physical check. An optimization answer is a sentence with units, and it should make sense: a width is positive, a count is a whole number, a price is not negative. A vertex that lands on an impossible input is a signal to re-read the setup.

Key takeaways

To maximize or minimize a quadratic quantity, find the vertex of its parabola. The optimizing input is $x = -\tfrac{b}{2a}$ ; the optimal value is what you get by substituting that input back (or the $k$ in vertex form $y = a(x - h)^2 + k$ ).
The sign of $a$ decides which extreme it is: $a < 0$ opens downward, so the vertex is a maximum; $a > 0$ opens upward, so the vertex is a minimum.
Keep the input and the value apart. One answers where the best occurs, the other answers how good it is, and a question wants one specific piece. They are rarely the same number.
Model in four steps: name the variable, write the quantity as a quadratic in it, then find the vertex. The fourth step is to answer exactly what is asked, with correct units and a sanity check.
Fixed perimeter: write one side as $x$ and the other as (half the perimeter) minus $x$ . That makes the area a downward parabola, largest when the rectangle is a square. Against a wall: with a fence of length $F$ , area $A = x(F - 2x)$ . Revenue: $R = (\text{price})(\text{quantity})$ with the quantity falling linearly as the price rises. Projectile: $h = -16t^2 + v_0 t + h_0$ peaks at its vertex.

Quadratic Optimization

Learning goals

Why the vertex is the best value

A method for optimization problems

Greatest area for a fixed amount of fence

Greatest revenue

A projectile’s highest point

Least cost and other minimums

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)