Word Problems on Quadratic Functions

$ y = P(x) \\[3ex] P(x) = -x^2 + 280x - 7500 \\[3ex] P(x) = -1 * x^2 + 280x - 7500 \\[3ex] (a.)\;\;To\;\;determine\;\;y-intercept \\[3ex] Set\;\;x = 0 \\[3ex] Solve\;\;for\;\;P(0) \\[3ex] P(x) = -1 * 0^2 + 280(0) - 7500 \\[3ex] P(0) = -1(0) + 0 - 7500 \\[3ex] P(0) = 0 + 0 - 7500 \\[3ex] P(0) = -7500 \\[3ex] y-intercept = (0, -7500) \\[3ex] $ According to the business model, if no Smart TV is sold on any work day in any week, the company will lose $7500.00


$ P(x) = -x^2 + 280x - 7500 \\[3ex] (b.)\;\;To\;\;determine\;\;x-intercept \\[3ex] Set\;\;P(x) = 0 \\[3ex] Solve\;\;for\;\;x \\[3ex] 0 = -x^2 + 280x - 7500 \\[3ex] -x^2 + 280x - 7500 = 0 \\[3ex] Divide\;\;both\;\;sides\;\;by\;\;-1 \\[3ex] x^2 - 280x + 7500 = 0 \\[3ex] x^2 - 250x - 30x + 7500 = 0 \\[3ex] x(x - 250) - 30(x - 250) = 0 \\[3ex] (x - 250)(x - 30) = 0 \\[3ex] x - 250 = 0 \;\;OR\;\; x - 30 = 0 \\[3ex] x = 250 \;\;OR\;\; x = 30 \\[3ex] x-intercepts\;\;are\;\; (30, 0) \;\;and\;\; (250, 0) \\[3ex] $ These are the zeros of the function.
According to the business model, the company will break even (make no profit or loss) if 30 TVs or 250 TVs are sold on any work day.
On any work day:
if less than 30 TVs are sold, the company will incur loss
if more than 250 TVs are sold, the company will also incur loss.

$ C(x) = cost\;\;function \\[3ex] R(x) = revenue\;\;function \\[3ex] P(x) = profit\;\;function \\[3ex] (a) \\[3ex] C(x) = 3.6x^2 + x \\[3ex] C(15) = 3.6(15)^2 + 15 \\[3ex] C(15) = 3.6(225) + 15 \\[3ex] C(15) = \$825.00 \\[3ex] (b) \\[3ex] R(x) = 73 * x = 73x \\[3ex] P(x) = R(x) - C(x) \\[3ex] P(x) = 73x - (3.6x^2 + x) \\[3ex] P(x) = 73x - 3.6x^2 - x \\[3ex] P(x) = -3.6x^2 + 72x \\[3ex] (c) \\[3ex] $ We can solve this question part using at least two approaches.
Vertex Method: the x-coordinate of the vertex gives the number of mats that should be sold to maximise the profit.
The y-coordinate of the vertex gives the maximum profit.

$ \underline{Vertex\;\;Method} \\[3ex] P(x) = -3.6x^2 + 72x \\[3ex] a = -3.6,\;\; b = 72,\;\; c = 0 \\[3ex] x-coordinate\;\;of\;\;vertex \\[3ex] = -\dfrac{b}{2a} \\[5ex] = -\dfrac{72}{2(-3.6)} \\[5ex] = -\dfrac{72}{-7.2} \\[5ex] = 10 \\[3ex] $
10 mats should be produced and sold to maximise profit.
The maximum profit is $360.00

$ \underline{Differential\;\;Calculus\;\;Approach} \\[3ex] P(x) = -3.6x^2 + 72x \\[3ex] P'(x) = -7.2x + 72 \\[3ex] Set\;\;P'(x) = 0 \;\;and\;\;solve\;\;for\;\;x \\[3ex] -7.2x + 72 = 0 \\[3ex] -7.2x = -72 \\[3ex] x = \dfrac{-72}{-7.2} \\[5ex] x = 10 \\[3ex] $ 10 mats should be produced and sold to maximise profit.

$ (a.) \\[3ex] x = -8p + 800 \\[3ex] R = xp \\[3ex] R(p) = p(-8p + 800) \\[3ex] R(p) = -8p^2 + 800p \\[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] -8p^2 + 800p \ge 0 \\[3ex] -8(p^2 - 100p) \ge 0 \\[3ex] p^2 - 100p \le \dfrac{0}{-8} \\[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 > 0 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 xx-coordinate of the revenue function.

$ R(p) = -8p^2 + 800p \\[3ex] Compare:\;\;R(p) = ap^2 + bp + c \\[3ex] a = -8 \\[3ex] b = 800 \\[3ex] x-coordinate\;\;of\;\;vertex = -\dfrac{b}{2a} \\[5ex] p = \dfrac{-800}{2(-8)} \\[5ex] = \dfrac{-800}{-16} \\[5ex] = 50 \\[3ex] $ The price that maximizes revenue is $50

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

$ R(p) = -8p^2 + 800p \\[3ex] p = \$50 \\[3ex] y-coordinate\;\;of\;\;vertex = R(50) \\[3ex] R(50) = -8(50)^2 + 800(50) \\[3ex] = -8(2500) + 40000 \\[3ex] = -20000 + 40000 \\[3ex] = 20000 \\[3ex] $ The maximum revenue is $20,000

(e.) The number of units sold at this price is the value of x for which R = $20000

$ vertex = (x, y) = (p, R) = (50, 20000) \\[3ex] R = xp \\[3ex] xp = R \\[3ex] x = \dfrac{R}{p} \\[5ex] x = \dfrac{20000}{50} \\[5ex] x = 400 \\[3ex] $ 400 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, 20000)



(g.) The price that the company should charge to earn at least $10,752

At least 10752 means ≥ 10752
They are asking us to determine the price range for which the revenue is greater than or equal to $10,752

$ -8p^2 + 800p \ge 10752 \\[3ex] -8p^2 + 800p - 10752 \ge 0 \\[3ex] -8(p^2 - 100p + 1344) \ge 0 \\[3ex] p^2 - 100p + 1344 \le \dfrac{0}{-8} \\[5ex] p^2 - 100p + 1344 \le 0 \\[3ex] (p - 16)(p - 84) \le 0 \\[3ex] Assume:\;\;(p - 16)(p - 84) = 0 \\[3ex] p - 16 = 0 \;\;\;OR\;\;\; p - 84 = 0 \\[3ex] p = 16 \;\;\;OR\;\;\; p = 84 \\[3ex] Test\;\;Intervals\;\;are: \\[3ex] p \lt 16 \\[3ex] 16 \le p \le 84 \\[3ex] p \gt 84 \\[3ex] $

Let:	p < 16 p = 0	16 ≤ p ≤ 84 p = 17	p > 84 p = 85
$p - 16$	−	+	+
$p - 84$	−	−	+
$(p - 16)(p - 84)$	+	−	+

Less than or equal to zero means nonnegative
Hence, the second test interval gives the solution of the inequality
The solution is: {p | 16 ≤ p ≤ 84}
The company should a price between a minimum of $16.00 and a maximum of $84.00

Notice the bold letter maximum in the question
The question is asking for the vertex

$ h(t) = -16t^2 + 28t + 2 \\[3ex] f(x) = ax^2 + bx + c \\[3ex] a = -16, b = 28, c = 2 \\[3ex] Vertex = \left(-\dfrac{b}{2a}, f\left(-\dfrac{b}{2a}\right)\right) \\[5ex] -\dfrac{b}{2a} = -\dfrac{28}{2(-16)} \\[5ex] -\dfrac{b}{2a} = \dfrac{-28}{-32} = \dfrac{7}{8} \\[5ex] f\left(-\dfrac{b}{2a}\right) = f\left(\dfrac{7}{8}\right) \\[5ex] But\:\: h(t) = -16t^2 + 28t + 2 \\[3ex] h\left(\dfrac{7}{8}\right) = -16\left(\dfrac{7}{8}\right)^2 + 28\left(\dfrac{7}{8}\right) + 2 \\[5ex] h\left(\dfrac{7}{8}\right) = -16\left(\dfrac{7}{8}\right)\left(\dfrac{7}{8}\right) + 7\left(\dfrac{7}{2}\right) + 2 \\[5ex] h\left(\dfrac{7}{8}\right) = -2\left(\dfrac{7}{1}\right)\left(\dfrac{7}{8}\right) + \dfrac{49}{2} + 2 \\[5ex] h\left(\dfrac{7}{8}\right) = -1\left(\dfrac{7}{1}\right)\left(\dfrac{7}{4}\right) + \dfrac{49}{2} + 2 \\[5ex] h\left(\dfrac{7}{8}\right) = -\dfrac{49}{4} + \dfrac{49}{2} + 2 \\[5ex] h\left(\dfrac{7}{8}\right) = -\dfrac{49}{4} + \dfrac{98}{4} + \dfrac{8}{4} \\[5ex] h\left(\dfrac{7}{8}\right) = \dfrac{-49 + 98 + 8}{4} \\[5ex] h\left(\dfrac{7}{8}\right) = \dfrac{57}{4} \\[5ex] h\left(\dfrac{7}{8}\right) = \dfrac{57}{4} \\[5ex] $
The time to reach the maximum height is $\dfrac{7}{8} = 0.875$ seconds

The maximum height is $\dfrac{57}{4} = 14.25$ feet

We can do this question in at least two ways.
You may use any of the methods to solve it.

First Method: Vertex Formula: this is because the profit is a quadratic function
For this question:
the number of bags that will give the maximum profit = $x$ coordinate of the vertex
the maximum profit = $y$ coordinate of the vertex

$ y = 10x - x^2 \\[3ex] y = -x^2 + 10x \\[3ex] y = ax^2 + bx + c \\[3ex] a = -1, b = 10 \\[3ex] x-coordinate\:\: of\:\: Vertex = -\dfrac{b}{2a} \\[5ex] = \dfrac{-10}{2(-1)} \\[5ex] = \dfrac{-10}{-2} \\[5ex] = 5 \\[3ex] $ The sale of 5 bags will give the maximum profit.
The maximum profit is 25 naira.



Second Method: Differential Calculus
One of the application of derivatives is in calculating maxima and minima of functions
$ y = 10x - x^2 \\[3ex] \dfrac{dy}{dx} = 10 - 2x \\[3ex] 10 - 2x = 0 \\[3ex] 10 = 2x \\[3ex] 2x = 10 \\[3ex] x = \dfrac{10}{2} \\[3ex] x = 5 \\[3ex] $ The sale of 5 bags will give the maximum profit.

This is an application of the vertex: ... values of $c$ will the weekly profit for this product be the largest?
This question requires us to find the $x-component$ of the vertex
The $x-component$ of the vertex is the price that will give the largest weekly profit
The $y-component$ of the vertex is the largest weekly profit

$ p(c) = 1,600c - 4c^2 \\[3ex] p(c) = -4c^2 + 1600c \\[3ex] Compare\;\;to\;\; ax^2 + bx + c \\[3ex] a = -4 \\[3ex] b = 1600 \\[3ex] c = 0 \\[3ex] $ This is not the actual value of $c$
It is the comparison to the standard form.
Anyway, we do not need the value of $c$ for this question.

$ x-component\;\;of\;\;vertex \\[3ex] = -\dfrac{b}{2a} \\[5ex] = -\dfrac{1600}{2(-4)} \\[5ex] = -\dfrac{1600}{-8} \\[5ex] = 200 \\[3ex] $
Based on the function: $p(c) = 1,600c - 4c^2$, the company needs to set the price at 200 cents for each unit of their new product in order to earn the largest weekly profit.

The question wants us to find the zeros of the function.

$ p(c) = 1,600c - 4c^2 \\[3ex] p(c) = -4c^2 + 1600c \\[3ex] Set\;\;p(c) = 0 \;\;and\;\;solve\;\;for\;\;c \\[3ex] 0 = -4c^2 + 1600c \\[3ex] -4c^2 + 1600c = 0 \\[3ex] -4c(c - 400) = 0 \\[3ex] -4c = 0 \;\;OR\;\; c - 400 = 0 \\[3ex] c = \dfrac{0}{-4} \;\;OR\;\; c = 0 + 400 \\[5ex] c = 0 \;\;OR\;\; c = 400 \\[3ex] $ Because the company cannot set the price at 0 cents (technically) because the employees most likely do not work for free, the company will break even (make no profit or loss) if the price of each unit of their new product is set at 400 cents.

Normal revenue for 1 week = $40 * 22 = \$880$
But, the store wants to make more money.
For every $\$1$ decrease in price, the store will sell $4$ more caps per week.
For $22 - 1 = 21$, the store will sell $40 + 4 = 44$ caps
The company will sell $4$ more caps for a $\$1$ decrease in price
Revenue would be $21 * 44 = \$924$
For $22 - 1(2) = 20$, the store will sell $40 + 4(2) = 40 + 8 = 48$ caps
The company will sell $8$ more caps for a $\$2$ decrease in price
Revenue would be $20 * 48 = \$960$
For $22 - 1(3) = 19$, the store will sell $40 + 4(3) = 40 + 12 = 52$ caps
The company will sell $12$ more caps for a $\$3$ decrease in price
Revenue would be $19 * 52 = \$988$

Student: That would exactly take more than a minute to solve.
ACT questions typically should be solved at at an average of $1$ question per minute.
Teacher: I expected that comment.
I did those steps to explain the question to you.
I was doing it the Arithmetic method.
Let us now do it algebraically.

So;
For each $\$x$ decrease in price, the company will sell $4x$ more
For $22 - 1(x) = 22 - x$, the store will sell $40 + 4(x) = 40 + 4x$ caps
Revenue would be $(22 - x) * (40 + 4x)$
Let Revenue = $R$

$ R = (22 - x)(40 + 4x) \\[3ex] = 880 + 88x - 40x - 4x^2 \\[3ex] = 880 + 48x - 4x^2 \\[3ex] = -4x^2 + 48x + 880 \\[3ex] R = -4x^2 + 48x + 880 \\[3ex] y = ax^2 + bx + c \\[3ex] a = -4, b = 48 \\[3ex] x-coordinate\:\: of\:\: vertex = -\dfrac{b}{2a} \\[5ex] = -\dfrac{48}{2(-4)} = \dfrac{-48}{-8} = 6 \\[5ex] $ The store has to decrease the price by $\$6$
This means that the store has to sell each cap for $22 - 6 = \$16$
By doing so, the store will sell $40 + 4(6) = 40 + 24 = 64$ caps
Revenue = $16 * 64 = \$1024$


Student: So, this is the maximum revenue the store can get?
Teacher: That is correct.
Student: What if the store sells each cap for $\$15$?
Teacher: Go ahead and find the revenue
Student: Okay...it will be
$15 * (64 + 4) = 15 * 68 = \$1020$
Interesting application of Quadratic Functions!

The price to maximise the revenue is the xx-coordinate of the vertex
The maximum revenue is the y-coordinate of the vertex.

$ (a) \\[3ex] R = -50x^2 + 4500x + 50 000 \\[3ex] a = -50 \\[3ex] b = 4500 \\[3ex] x-coordinate\;\;of\;\;the\;\;vertex \\[3ex] = -\dfrac{b}{2a} \\[5ex] = - \dfrac{4500}{2(-50)} \\[5ex] = \dfrac{-4500}{-100} \\[5ex] = 45 \\[3ex] $ However, we noticed from the graph that 10 units of x was cut off
To account for those units due to symmetry, we should add 10 to the value of x
Therefore, the price to maximise revenue = 45 + 10 = $55.00

(b)
To find the value of the intercept of the parabola with the vertical axis means that we should find the y-intercept
To find the y-intercept, set the value of x to be zero and solve for y

$ R = -50x^2 + 4500x + 50 000 \\[3ex] set\;\;x = 0 \\[3ex] \implies \\[3ex] R = -50(0)^2 + 4500(0) + 50 000 \\[3ex] R = 0 + 0 + 50000 \\[3ex] R = \$50000.00 \\[3ex] $ The value of the intercept of the parabola with the vertical axis is $50 000.00