Solved Examples on Quadratic Models

The Forms of a Quadratic Function are:

(1.) Standard Form/General Form
Note that the standard form is written in descending order of x

$ y = ax^2 + bx + c \\[3ex] OR \\[3ex] f(x) = ax^2 + bx + c \\[3ex] Vertex = \left[-\dfrac{b}{2a}, f\left(-\dfrac{b}{2a}\right)\right] \\[5ex] $ (2.) Vertex Form

$ y = a(x - h)^2 + k \\[3ex] OR \\[3ex] f(x) = a(x - h)^2 + k \\[3ex] Vertex = (h, k) \\[3ex] h = x-coordinate\;\;of\;\;the\;\;vertex \\[3ex] k = y-coordinate\;\;of\;\;the\;\;vertex \\[3ex] $ (3.) Extended Vertex Form

$ f(x) = a(x - h)^2 + k ...Vertex\;\;Form \\[3ex] f(x) = a\left(x + \dfrac{b}{2a}\right)^2 + \dfrac{4ac - b^2}{4a}...Extended\;\;Vertex\;\;Form \\[5ex] Vertex = (h, k) \\[3ex] \implies \\[3ex] -h = \dfrac{b}{2a} \\[5ex] h = -\dfrac{b}{2a} \\[5ex] k = \dfrac{4ac - b^2}{4a} \\[5ex] \implies \\[3ex] \boldsymbol{Vertex = (h, k) = \left(-\dfrac{b}{2a}, \dfrac{4ac - b^2}{4a}\right)} \\[5ex] $ Attempt all questions.
Show all work.

(1.) The price p (in dollars) and the quantity x sold of a certain product satisfy the demand equation x = −9p + 900

(a.) Find a model that expresses the revenue R as a function of p.
(Remember R = xp.)

(b.) What is the domain of​ R?
Assume R is nonnegative.

​(c.) What price p maximizes​ revenue?
Simplify your answer. Type an integer or decimal.

​(d.) What is the maximum​ revenue?
Simplify your answer. Type an integer or decimal.

(e.) How many units are sold at this​ price?
Simplify your answer. Type an integer or decimal.

(f.) Graph R

Number 1

​(g.) What price should the company charge to earn at least $10,179 in​ revenue?
Simplify your answer. Type an integer or decimal.


$ (a.) \\[3ex] R(p) = x * p \\[3ex] = p * x \\[3ex] = p(-9p + 900) \\[3ex] = -9p^2 + 900p \\[3ex] $ (b.) The revenue cannot be negative.
The revenue can be zero (no revenue)
The revenue can be positive
Hence, the revenue must be nonnegative
The domain of R includes all the real number prices such that the revenue is nonnegative

$ R(p) \ge 0 \\[3ex] -9p^2 + 900p \ge 0 \\[3ex] -9(p^2 - 100p) \ge 0 \\[3ex] p^2 - 100p \le \dfrac{0}{-9} \\[5ex] p^2 - 100p \le 0 \\[3ex] p(p - 100) \le 0 \\[3ex] Assume:\;\;p(p - 100) = 0 \\[3ex] p = 0 \;\;\;OR\;\;\; p - 100 = 0 \\[3ex] p = 0 \;\;\;OR\;\;\; p = 100 \\[3ex] Test\;\;Intervals\;\;are: \\[3ex] p \lt 0 \\[3ex] 0 \le p \le 100 \\[3ex] p \gt 100 \\[3ex] $

Let: 		p < 0
p = −1 		0 ≤ p ≤ 100
p = 1 		p > 100
p = 101
$p$ − + +
$p - 100$ − − +
$p(p - 100)$ + − +

Less than or equal to zero means nonnegative
Hence, the second test interval gives the solution of the inequality
Domain, D = {p | 0 ≤ p ≤ 100}

(c.) The price that maximizes revenue is the x-coordinate of the vertex of the revenue function.

$ R(p) = -9p^2 + 900p \\[3ex] Compare:\;\;R(p) = ap^2 + bp + c \\[3ex] a = -9 \\[3ex] b = 900 \\[3ex] x-coordinate\;\;of\;\;vertex = -\dfrac{b}{2a} \\[5ex] p = \dfrac{-900}{2(-9)} \\[5ex] = \dfrac{-900}{-18} \\[5ex] = 50 \\[3ex] $ The price that maximizes revenue is $50

(d.) The maximum revenue is the y-coordinate of the vertex of the revenue function.

$ R(p) = -9p^2 + 900p \\[3ex] p = \$50 \\[3ex] y-coordinate\;\;of\;\;vertex = R(50) \\[3ex] R(50) = -9(50)^2 + 900(50) \\[3ex] = -9(2500) + 45000 \\[3ex] = -22500 + 45000 \\[3ex] = 22500 \\[3ex] $ The maximum revenue is $22,500

(e.) The number of units sold at this price is the value of x for which R = $22500

$ vertex = (x, y) = (p, R) = (50, 22500) \\[3ex] R = xp \\[3ex] xp = R \\[3ex] x = \dfrac{R}{p} \\[5ex] x = \dfrac{22500}{50} \\[5ex] x = 450 \\[3ex] $ 450 units were sold at the price that gave the maximum revenue.

(f.) The graph of the revenue versus price is shown below.
Keep in mind that the vertex is (p, R) = (50, 22500)
Options (A.) and (D.) are similar.
However, option (A.) is the graph of R versus p while option (D.) is the graph of p versus R
The revenue, R is the y-axis
The price that maximizes the revenue, p is the x-axis
So, the correct graph should be the graph of R versus p: option (A.)

Number 1

(g.) The price that the company should charge to earn at least $10,179

At least 10179 means ≥ 10179
They are asking us to determine the price range for which the revenue is greater than or equal to $10,179

$ -9p^2 + 900p \ge 10179 \\[3ex] -9p^2 + 900p - 10179 \ge 0 \\[3ex] -9(p^2 - 100p + 1131) \ge 0 \\[3ex] p^2 - 100p + 1131 \le \dfrac{0}{-9} \\[5ex] p^2 - 100p + 1131 \le 0 \\[3ex] (p - 13)(p - 87) \le 0 \\[3ex] Assume:\;\;(p - 13)(p - 87) = 0 \\[3ex] p - 13 = 0 \;\;\;OR\;\;\; p - 87 = 0 \\[3ex] p = 13 \;\;\;OR\;\;\; p = 87 \\[3ex] Test\;\;Intervals\;\;are: \\[3ex] p \lt 13 \\[3ex] 13 \le p \le 87 \\[3ex] p \gt 87 \\[3ex] $

Let: 		p < 13
p = 0 		13 ≤ p ≤ 87
p = 17 		p > 87
p = 87
$p - 13$ − + +
$p - 87$ − − +
$(p - 13)(p - 87)$ + − +

Less than or equal to zero means nonnegative
Hence, the second test interval gives the solution of the inequality
The solution is: {p | 13 ≤ p ≤ 87}
The company should a price between a minimum of $13.00 and a maximum of $87.00
(2.) Jude has 9500 meters of​ fencing, and wants to enclose a rectangular plot that borders on a river.
Number 2
If Jude does not fence the side along the​ river, what is the largest area that can be​ enclosed?


Let the:
area = A
length = x

$ A = x(9500 - 2x) \\[3ex] A = 9500x - 2x^2 \\[3ex] A = -2x^2 + 9500x \\[3ex] $ The largest area is the y-coordinate of the vertex of the area function.

$ A = -2x^2 + 9500x \\[3ex] Compare:\;\;A = ax^2 + bx + c \\[3ex] a = -2 \\[3ex] b = 9500 \\[3ex] c = 0 \\[3ex] A = \dfrac{4ac - b^2}{4a} \\[5ex] = \dfrac{4(-2)(0) - (9500)^2}{4(-2)} \\[5ex] = \dfrac{0 - 90250000}{-8} \\[5ex] = \dfrac{-90250000}{-8} \\[5ex] = 11281250 \\[3ex] $ The largest area is 11281250 square meters.
(3.) Phoebe has available 2800 yards of fencing and wishes to enclose a rectangular area.
​(a.) Express the area A of the rectangle as a function of the width W of the rectangle.
​(b.) For what value of W is the area​ largest?
​(c.) What is the maximum​ area?


Let the:
Length of the rectangle = L
Perimeter of the rectangle = P

$ \underline{Formulas} \\[3ex] P = 2(L + W)...eqn.(1) \\[3ex] A = L * W...eqn.(2) \\[3ex] P = 2800\;yards...Given \\[3ex] \implies \\[3ex] 2(L + W) = 2800 \\[3ex] L + W = \dfrac{2800}{2} \\[5ex] L + W = 1400 \\[3ex] L = 1400 - W ...eqn.(3) \\[3ex] $ To express the area as a function of the width, we had to express the length in term of the width.
Then, we have to substitute that value of the length in the formula for the area.

$ (a.) \\[3ex] Substitute\;\;eqn.(3)\;\;into\;\;eqn.(2) \\[3ex] A = (1400 - W) * W \\[3ex] A = W(1400 - W) \\[3ex] A = 1400W - W^2 \\[3ex] A = -W^2 + 1400W \\[3ex] $ (b.) The value of the width for which the area is the largest is the x-coordinate of the vertex of the area function.

$ A = -W^2 + 1400W \\[3ex] Compare:\;\;A = aW^2 + bW + c \\[3ex] a = -1 \\[3ex] b = 1400 \\[3ex] x-coordinate\;\;of\;\;vertex = -\dfrac{b}{2a} \\[5ex] W = \dfrac{-1400}{2(-1)} \\[5ex] = \dfrac{-1400}{-2} \\[5ex] = 700 \\[3ex] $ The area of the rectangle is largest when the width is 700 yards.

(c.) The maximum area is the y-coordinate of the vertex of the area function.

$ A(700) = -(700)^2 + 1400(700) \\[3ex] = -490000 + 980000 \\[3ex] = 490000 \\[3ex] $ The maximum area is 490,000 square yards.
(4.) The price p (in dollars) and the quantity x sold of a certain product satisfy the demand equation x = −6p + 300

(a.) Find a model that expresses the revenue R as a function of p.
(Remember R = xp.)

(b.) What is the domain of​ R?
Assume R is nonnegative.

​(c.) What price p maximizes​ revenue?
Simplify your answer. Type an integer or decimal.

​(d.) What is the maximum​ revenue?
Simplify your answer. Type an integer or decimal.

(e.) How many units are sold at this​ price?
Simplify your answer. Type an integer or decimal.

(f.) Graph R

Number 4

​(g.) What price should the company charge to earn at least $2736 in​ revenue?
Simplify your answer. Type an integer or decimal.


$ (a.) \\[3ex] R(p) = x * p \\[3ex] = p * x \\[3ex] = p(-6p + 300) \\[3ex] = -6p^2 + 300p \\[3ex] $ (b.) The revenue cannot be negative.
The revenue can be zero (no revenue)
The revenue can be positive
Hence, the revenue must be nonnegative
The domain of R includes all the real number prices such that the revenue is nonnegative

$ R(p) \ge 0 \\[3ex] -6p^2 + 300p \ge 0 \\[3ex] -6(p^2 - 50p) \ge 0 \\[3ex] p^2 - 50p \le \dfrac{0}{-6} \\[5ex] p^2 - 50p \le 0 \\[3ex] p(p - 50) \le 0 \\[3ex] Assume:\;\;p(p - 50) = 0 \\[3ex] p = 0 \;\;\;OR\;\;\; p - 50 = 0 \\[3ex] p = 0 \;\;\;OR\;\;\; p = 50 \\[3ex] Test\;\;Intervals\;\;are: \\[3ex] p \lt 0 \\[3ex] 0 \le p \le 50 \\[3ex] p \gt 50 \\[3ex] $

Let: 		p < 0
p = −1 		0 ≤ p ≤ 50
p = 1 		p > 50
p = 51
$p$ − + +
$p - 50$ − − +
$p(p - 50)$ + − +

Less than or equal to zero means nonnegative
Hence, the second test interval gives the solution of the inequality
Domain, D = {p | 0 ≤ p ≤ 50}

(c.) The price that maximizes revenue is the x-coordinate of the vertex of the revenue function.

$ R(p) = -6p^2 + 300p \\[3ex] Compare:\;\;R(p) = ap^2 + bp + c \\[3ex] a = -6 \\[3ex] b = 300 \\[3ex] x-coordinate\;\;of\;\;vertex = -\dfrac{b}{2a} \\[5ex] p = \dfrac{-300}{2(-6)} \\[5ex] = \dfrac{-300}{-12} \\[5ex] = 25 \\[3ex] $ The price that maximizes revenue is $25

(d.) The maximum revenue is the y-coordinate of the vertex of the revenue function.

$ R(p) = -6p^2 + 300p \\[3ex] p = \$25 \\[3ex] y-coordinate\;\;of\;\;vertex = R(25) \\[3ex] R(25) = -6(25)^2 + 300(25) \\[3ex] = -6(625) + 7500 \\[3ex] = -3750 + 7500 \\[3ex] = 3750 \\[3ex] $ The maximum revenue is $3750

(e.) The number of units sold at this price is the value of x for which R = $3750

$ vertex = (x, y) = (p, R) = (25, 3750) \\[3ex] R = xp \\[3ex] xp = R \\[3ex] x = \dfrac{R}{p} \\[5ex] x = \dfrac{3750}{25} \\[5ex] x = 150 \\[3ex] $ 150 units were sold at the price that gave the maximum revenue.

(f.) The graph of the revenue versus price is shown below.
Keep in mind that the vertex is (p, R) = (25, 3750)
Options (B.) and (C.) are similar.
However, option (B.) is the graph of R versus p while option (C.) is the graph of p versus R
The revenue, R is the y-axis
The price that maximizes the revenue, p is the x-axis
So, the correct graph should be the graph of R versus p: option (B.)

Number 4

(g.) The price that the company should charge to earn at least $2736

At least 2736 means ≥ 2736
They are asking us to determine the price range for which the revenue is greater than or equal to $2736

$ -6p^2 + 300p \ge 2736 \\[3ex] -6p^2 + 300p - 2736 \ge 0 \\[3ex] -6(p^2 - 50p + 456) \ge 0 \\[3ex] p^2 - 50p + 456 \le \dfrac{0}{-6} \\[5ex] p^2 - 50p + 456 \le 0 \\[3ex] (p - 12)(p - 38) \le 0 \\[3ex] Assume:\;\;(p - 12)(p - 36) = 0 \\[3ex] p - 12 = 0 \;\;\;OR\;\;\; p - 36 = 0 \\[3ex] p = 12 \;\;\;OR\;\;\; p = 36 \\[3ex] Test\;\;Intervals\;\;are: \\[3ex] p \lt 12 \\[3ex] 12 \le p \le 38 \\[3ex] p \gt 38 \\[3ex] $

Let: 		p < 12
p = 0 		12 ≤ p ≤ 38
p = 16 		p > 38
p = 39
$p - 12$ − + +
$p - 38$ − − +
$(p - 12)(p - 38)$ + − +

Less than or equal to zero means nonnegative
Hence, the second test interval gives the solution of the inequality
The solution is: {p | 12 ≤ p ≤ 38}
The company should a price between a minimum of $12.00 and a maximum of $38.00
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(11.) A suspension bridge with weight uniformly distributed along its length has twin towers that extend 50 meters above the road surface and are 800 meters apart.
The cables are parabolic in shape and are suspended from the tops of the towers.
The cables touch the road surface at the center of the bridge.
Find the height of the cables at a point 200 meters from the center.
Assume that the road is​ level.


Let us make a diagrammatic representation of this information.
Number 11
Let the height of the cable at a point 200 meters from the center = h

$ y = ax^2...Quadratic\;\;Function...parabola \\[3ex] For\;\;(400, 50) \\[3ex] x = 400 \\[3ex] y = 50 \\[3ex] \implies \\[3ex] 50 = a(400)^2 \\[3ex] a(400)^2 = 50 \\[3ex] a = \dfrac{50}{400^2} \\[5ex] a = \dfrac{50}{400(400)} \\[5ex] a = \dfrac{5}{16000} \\[5ex] For\;\;(200, h) \\[3ex] x = 200 \\[3ex] y = h \\[3ex] h = ax^2 \\[3ex] h = \dfrac{5}{16000} * (200)^2 \\[5ex] = \dfrac{5}{16000} * \dfrac{200 * 200}{1} \\[5ex] = \dfrac{50}{4} \\[5ex] = \dfrac{25}{2} \\[5ex] = 12.5 \\[3ex] $ The height of the cable at a point 200 meters from the center is 12.5 meters.
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