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Quadratic Optimization: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A best value and where it happens

For the quantity $y=3x^2-24x+50$ , with $x$ any real number, state whether it has a maximum or a minimum. Give the input where that happens and the value there.
1. Hint 1
  The sign of the leading coefficient decides whether the vertex is a lowest or a highest point.
2. Hint 2
  Locate the vertex input with $-\frac{b}{2a}$ , then substitute that input back to get the value there.
Answer

A minimum; it occurs at $x=4$ , and the value there is $2$ .

Full solution

The leading coefficient is $a=3$ , which is positive, so the parabola opens upward and its vertex is the lowest point.
The quantity therefore has a minimum.

The minimum occurs at the vertex input, read with $a=3$ and $b=-24$ .
$x=-\frac{-24}{2(3)}$

This is $x=4$ .
Substitute that input back into the quantity.
$y=3(4)^2-24(4)+50$

This is $48-96+50=2$ , so the minimum value is $2$ .
The input $4$ says where the minimum happens, and the value $2$ says how small the quantity gets.

Answer

A minimum; it occurs at $x=4$ , and the value there is $2$ .

Key idea

A positive leading coefficient makes the vertex a minimum, and its two coordinates answer two different questions.
Problem 1: Mark complete
Problem 2 A quantity written as a product

A quantity is given as a product, $y=(x+8)(12-3x)$ , where $x$ may be any real number. Find the input at which $y$ is greatest, and that greatest value.
1. Hint 1
  A product of two linear expressions is a quadratic, so its best value, greatest or least, sits at the vertex of its parabola.
2. Hint 2
  Multiply out to read off $a$ and $b$ , then use $-\frac{b}{2a}$ for the input. The factored form is the quicker one to evaluate there.
Answer

$y$ is greatest at $x=-2$ ; the greatest value is $108$ .

Full solution

Multiply the two factors out.
The two $x$ terms collect as $12x-24x=-12x$ .
$y=-3x^2-12x+96$

Read $a=-3$ and $b=-12$ , and take the vertex input.
$x=-\frac{-12}{2(-3)}$

This is $x=-2$ , and since $a=-3$ is negative the parabola opens downward, so that input is where $y$ is greatest.

Evaluating is quicker in the factored form: at $x=-2$ the factors are $6$ and $18$ , so the greatest value is $108$ .

Answer

$y$ is greatest at $x=-2$ ; the greatest value is $108$ .

Key idea

Expanding a factored quadratic exposes the $a$ and $b$ that locate its vertex, while the factored form is the quicker place to evaluate there.
Problem 2: Mark complete
Problem 3 The least value of a sum of two squares

For real $x$ , find the least possible value of $(x-1)^2+(x-5)^2$ .
1. Hint 1
  Combine the two squares into one quadratic.
2. Hint 2
  Write the total as one squared expression plus a constant to identify its smallest value.
Answer

$8$ .

Full solution

Expanding both squares and collecting like terms, the total is
$2x^2-12x+26$

Complete the square.
$2x^2-12x+26=2(x-3)^2+8$

The squared term is zero or positive and vanishes at $x=3$ .
The least value is $8$ , checked directly by
$2^2+(-2)^2=8.$

Answer

$8$ .

Key idea

Adding two squares of linear expressions gives one quadratic, whose least value comes from completing its square.
Problem 3: Mark complete
Problem 4 A required long side

A rectangle has perimeter $26$ meters, and its length must be at least $9$ meters. Its width and length are positive real numbers. Find the greatest possible area and the dimensions that give it.
1. Hint 1
  Express one dimension in terms of the other using the perimeter.
2. Hint 2
  Find the vertex of the area quadratic and check whether that input meets the length restriction.
Answer

Greatest area $36$ square meters; length $9$ meters and width $4$ meters.

Full solution

Let the width be $w$ .
The length is $13-w$ , and the restriction gives $0<w\le4$ .
The area is
$A=13w-w^2$

Its vertex is at $w=13/2$ , outside this interval.
Completing the square gives
$A=-\left(w-\frac{13}{2}\right)^2+\frac{169}{4}$

The value is greatest at the allowed width closest to $13/2$ , namely $w=4$ .
The length is $9$ meters and area is $4(9)=36$ square meters.
Their perimeter is $2(4+9)=26$ meters.

Answer

Greatest area $36$ square meters; length $9$ meters and width $4$ meters.

Key idea

A dimension restriction can move the best permitted rectangle away from the unrestricted square.
Problem 4: Mark complete
Problem 5 Admission and refreshments

An event charges $p$ dollars for admission and a required $3$ dollars for refreshments to each attendee. Predicted attendance is $40-2p$ . The price $p$ may be any multiple of $0.5$ from $0$ through $20$ . Find the admission price that maximizes total receipts and the greatest total receipts.
1. Hint 1
  Each attendee pays the admission price plus the refreshment charge.
2. Hint 2
  Multiply the total charge by attendance, find the vertex input, and check that it is a permitted price.
Answer

Admission price $8.50$ dollars, or $8.5$ ; greatest total receipts $264.50$ dollars.

Full solution

Total receipts in dollars are
$R=(p+3)(40-2p)$

Expanding gives
$R=-2p^2+34p+120$

The leading coefficient is negative, so the vertex gives a maximum.
$p=-\frac{34}{2(-2)}$

This is $8.50$ , a permitted price.
Attendance is $23$ , and each attendee pays $11.50$ dollars.
Total receipts are $11.50(23)=264.50$ dollars.

Answer

Admission price $8.50$ dollars, or $8.5$ ; greatest total receipts $264.50$ dollars.

Key idea

When each buyer pays an additional fixed charge, total receipts use that charge as well as the adjustable price.
Problem 5: Mark complete
Problem 6 A delayed recording

A ball has height $h=-16t^2+72t+12$ feet at time $t$ seconds after launch. A camera begins recording $1$ second after launch and stops $3.5$ seconds after launch. How long after recording begins does the camera capture the greatest height, and what is that height?
1. Hint 1
  Find the time of the vertex using the launch clock.
2. Hint 2
  Check that this time falls inside the recording period, then convert to the camera’s elapsed time and evaluate the height.
Answer

$1.25$ seconds after recording begins; greatest height $93$ feet.

Full solution

The downward height quadratic peaks at
$t=-\frac{72}{2(-16)}$

This is $2.25$ seconds after launch, within the recording period.
The camera has then been recording for $2.25-1=1.25$ seconds.
Substitute $t=2.25$ , the time on the launch clock, into the model, where $-16(2.25)^2=-81$ and $72(2.25)=162$ .
$h=-81+162+12$

The height is $93$ feet.
Both times around the peak within the recording interval have smaller heights, so the camera captures the actual flight maximum.

Answer

$1.25$ seconds after recording begins; greatest height $93$ feet.

Key idea

Find the optimizing time in the model’s clock before converting it to another elapsed-time reference.
Problem 6: Mark complete
Problem 7 Two operating costs

Two machines must use the same whole-number setting $x\ge0$ . Their costs in dollars are $C_1=x^2-8x+30$ and $C_2=2x^2-4x+18$ . Find the setting that minimizes their combined cost and the least combined cost.
1. Hint 1
  Combine the costs before optimizing, since both depend on the same setting.
2. Hint 2
  Find the vertex of the total and verify that its setting is allowed.
Answer

Setting $x=2$ ; least combined cost $36$ dollars.

Full solution

Add the two costs.
$C_1+C_2=3x^2-12x+48$

The leading coefficient is positive, so the vertex is a minimum at
$x=-\frac{-12}{2(3)}$

This gives $x=2$ , an allowed whole number.
The separate costs are $18$ dollars and $18$ dollars, giving $36$ dollars in total.
The total also equals $3(x-2)^2+36$ , confirming the minimum.

Answer

Setting $x=2$ ; least combined cost $36$ dollars.

Key idea

Optimize a combined cost as a single quantity when its parts must share one input.
Problem 7: Mark complete
Problem 8 A pricing claim

A shop sells $30-p$ items at a price of $p$ dollars each, where $0\le p\le30$ and $p$ is a whole number. Each item costs the shop $6$ dollars. A student says the price that gives the greatest revenue also gives the greatest profit. Decide whether this is correct by finding both prices.
1. Hint 1
  Revenue uses the full price per item, while profit uses price minus cost.
2. Hint 2
  Build and optimize the two products separately.
Answer

False; revenue is greatest at $15$ dollars, while profit is greatest at $18$ dollars.

Full solution

Revenue is
$R=p(30-p)$

This downward quadratic has its vertex at $p=15$ .
Each item sold earns $p-6$ dollars once the shop's cost is taken off, and $30-p$ items sell, so profit is the product of those two.
$P=(p-6)(30-p)$

Expanding gives $P=-p^2+36p-180$ , whose vertex is at $p=18$ .
Both are permitted whole-number prices.
Since the maximizing prices differ, the claim is false.

Answer

False; revenue is greatest at $15$ dollars, while profit is greatest at $18$ dollars.

Key idea

A cost per item changes the quadratic for profit and can change its maximizing price.
Problem 8: Mark complete
Problem 9 A higher launch

Two balls start at or above ground level with the same positive initial upward speed. One is launched from a platform $7$ feet higher than the other. Both follow $h=-16t^2+v_0t+h_0$ until they reach the ground. A student says they reach their greatest heights at the same elapsed time, and the higher launch gives a greatest height exactly $7$ feet greater. Is the claim correct? Explain.
1. Hint 1
  Only the constant term differs in the two height models.
2. Hint 2
  Compare the two vertex times using only the leading and linear coefficients.
Answer

Yes; the peak times agree and the greatest heights differ by $7$ feet.

Full solution

The shared vertex time is
$t=\frac{v_0}{32}$

It is positive because $v_0>0$ , and it occurs during each flight.
The other model has the same leading and linear coefficients, so its peak time agrees.

At that time, as at every common modeled time, the higher ball’s height is the lower ball’s height plus $7$ .
Their greatest heights therefore differ by $7$ feet.

Answer

Yes; the peak times agree and the greatest heights differ by $7$ feet.

Key idea

Changing the starting height adds the same amount to the peak height while leaving the peak time unchanged.
Problem 9: Mark complete
Problem 10 A rounded setting

A machine’s score is $Q=10-(x-2.5)^2$ , and $x$ must be a whole number from $0$ through $5$ . A student rounds the vertex input down and says that $x=2$ is the unique best setting. Is the claim correct? Give every best setting and the greatest permitted score.
1. Hint 1
  A quadratic's value depends only on how far the input is from the vertex input.
2. Hint 2
  Compare the distances of the two neighboring whole numbers from the vertex input.
Answer

False; $x=2$ and $x=3$ both give the greatest permitted score $9.75$ .

Full solution

The unrestricted vertex input is $2.5$ .
Both $2$ and $3$ are $0.5$ from it, so each gives
$Q=10-(0.5)^2$

This equals $9.75$ .
Every other permitted whole number lies farther from $2.5$ and subtracts a larger square.
Thus there are two best settings.

Answer

False; $x=2$ and $x=3$ both give the greatest permitted score $9.75$ .

Key idea

Two permitted inputs equally close to the vertex can tie for the best quadratic value.
Problem 10: Mark complete

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