solve the following quadratic word problems
3
3)
a) what is the width and what is the length of a rectangle
with an area of 405cm² if the length is 3 less than 2 times
the width?
b) a soccer ball being kicked off the roof of a building, is
modelled by
h(t) = -4(t - 7)² + 256 where t is measured in
seconds and h is measured in metres
i. what is the height of the soccer ball at 4 seconds?
ii. when does the soccer ball hit the ground?
iii. when does the soccer ball reach its max height, and
what is that height?
c) a robotics company is trying to maximize their sales of
robotic giraffes. the cost model, in dollars, to produce x
giraffes is
c(x) = 72x + 590 and their possible revenue is modelled
by r(x) = -2x² + 200x.
i. how many giraffes should they sell to reach maximum
profit?
ii. what is the maximum possible profit?
iii. what should the price per giraffe be to maximize
profit?

Question

Accepted Answer

### a)
# Explanation:
## Step1: Define variables
Let the width of the rectangle be $w$ cm. Then the length $l=(2w - 3)$ cm.
The area of a rectangle $A=l\times w$. Given $A = 405$ $cm^{2}$, so $w(2w - 3)=405$.
## Step2: Expand and solve the quadratic equation
Expand $w(2w - 3)=405$ to get $2w^{2}-3w - 405=0$.
We can use the quadratic formula $w=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $ax^{2}+bx + c = 0$. Here $a = 2$, $b=-3$, $c=-405$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(-3)^{2}-4\times2\times(-405)=9 + 3240=3249$.
Then $w=\frac{3\pm\sqrt{3249}}{4}=\frac{3\pm57}{4}$.
We have two solutions for $w$: $w_{1}=\frac{3 + 57}{4}=\frac{60}{4}=15$ and $w_{2}=\frac{3-57}{4}=\frac{-54}{4}=- 13.5$. Since width cannot be negative, $w = 15$ cm.
## Step3: Find the length
If $w = 15$ cm, then $l=2w-3=2\times15-3=27$ cm.

# Answer:
The width of the rectangle is $15$ cm and the length is $27$ cm.

### b)
#### i.
# Explanation:
## Step1: Substitute $t = 4$ into the height function
Given $h(t)=-4(t - 7)^{2}+256$. Substitute $t = 4$ into the function:
$h(4)=-4(4 - 7)^{2}+256$.
## Step2: Simplify the expression
First, calculate $(4 - 7)^{2}=(-3)^{2}=9$. Then $h(4)=-4\times9+256=-36 + 256=220$

# Answer:
The height of the soccer - ball at $4$ seconds is $220$ metres.

#### ii.
# Explanation:
## Step1: Set $h(t)=0$
Set $h(t)=-4(t - 7)^{2}+256 = 0$.
Then $-4(t - 7)^{2}=-256$.
## Step2: Solve for $t$
Divide both sides by $-4$: $(t - 7)^{2}=64$.
Take the square root of both sides: $t-7=\pm8$.
Case 1: $t-7 = 8$, then $t=8 + 7=15$.
Case 2: $t-7=-8$, then $t=-8 + 7=-1$. Since time $t\geq0$, we discard $t=-1$.

# Answer:
The soccer - ball hits the ground at $t = 15$ seconds.

#### iii.
# Explanation:
## Step1: Analyze the vertex form of the quadratic function
The function $h(t)=-4(t - 7)^{2}+256$ is in the vertex form $y=a(x - h)^{2}+k$, where $(h,k)$ is the vertex of the parabola.
For a quadratic function $y = a(x - h)^{2}+k$, if $a<0$ (here $a=-4<0$), the parabola opens downwards and the vertex $(h,k)$ is the maximum point.

# Answer:
The soccer - ball reaches its maximum height at $t = 7$ seconds and the maximum height is $256$ metres.

### c)
#### i.
# Explanation:
## Step1: Define the profit function
Profit $P(x)=R(x)-C(x)$. Given $R(x)=-2x^{2}+200x$ and $C(x)=72x + 590$.
$P(x)=-2x^{2}+200x-(72x + 590)=-2x^{2}+200x-72x - 590=-2x^{2}+128x - 590$.
## Step2: Find the vertex of the quadratic function
For a quadratic function $y = ax^{2}+bx + c$ (here $a=-2$, $b = 128$), the $x$ - coordinate of the vertex is $x=-\frac{b}{2a}$.
$x=-\frac{128}{2\times(-2)}=\frac{-128}{-4}=32$

# Answer:
They should sell $32$ giraffes to reach maximum profit.

#### ii.
# Explanation:
## Step1: Substitute $x = 32$ into the profit function
Substitute $x = 32$ into $P(x)=-2x^{2}+128x - 590$.
$P(32)=-2\times32^{2}+128\times32-590$.
First, $32^{2}=1024$, then $-2\times1024=-2048$, $128\times32 = 4096$.
$P(32)=-2048+4096-590=1458$

# Answer:
The maximum possible profit is $1458$ dollars.

#### iii.
# Explanation:
## Step1: Find the price per giraffe
The revenue function $R(x)=-2x^{2}+200x$. The price per unit $p=\frac{R(x)}{x}$ (for $x
eq0$).
Since $x = 32$ (from part i) and $R(x)=-2x^{2}+200x$, when $x = 32$, $R(32)=-2\times32^{2}+200\times32=-2048 + 6400=4352$.
The price per giraffe $p=\frac{R(32)}{32}=\frac{4352}{32}=136$

# Answer:
The price per giraffe should be $136$ dollars to maximize profit.

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

a)

Step1: Define variables

Step2: Expand and solve the quadratic equation

Step3: Find the length

Step1: Substitute \(t = 4\) into the height function

Step2: Simplify the expression

Step1: Set \(h(t)=0\)

Step2: Solve for \(t\)

Answer:

b)

i.