Question

i. exploring quadratic functions
topic practice
a. write a quadratic function in standard form that represents
each area as a function of the width. remember to define
your variables.

1. a builder is designing a rectangular parking lot. she has 300 feet of fencing to enclose the parking lot around three sides.
2. olivia is enclosing a new rectangular flower garden with a rabbit garden fence. she has 40 meters of fencing.
3. michael is building a rectangular sandbox for the community park. the materials available limit the perimeter of the sandbox to at most 100 feet.
4. daniela is designing a rectangular quilt. she has 16 feet of piping to finish the quilt around three sides.
5. gabriela is making a rectangular vegetable garden alongside her home. she has 24 meters of fencing to enclose the garden around the three open sides.
6. javier is building a rectangular ice rink for the community park. the materials available limit the perimeter of the ice rink to at most 250 meters.

Explanation:

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Problem 1

Step1: Define variables

Let $x$ = width of the lot (ft), $y$ = length of the lot (ft). Fencing covers 2 widths + 1 length: $2x + y = 300$.

Step2: Solve for length

$y = 300 - 2x$

Step3: Write area function

Area $A(x) = x \cdot y = x(300 - 2x) = -2x^2 + 300x$

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Problem 2

Step1: Define variables

Let $x$ = width of the garden (m), $y$ = length of the garden (m). Fencing covers all 4 sides: $2x + 2y = 40$.

Step2: Simplify and solve for length

$x + y = 20 \implies y = 20 - x$

Step3: Write area function

Area $A(x) = x \cdot y = x(20 - x) = -x^2 + 20x$

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Problem 3

Step1: Define variables

Let $x$ = width of the sandbox (ft), $y$ = length of the sandbox (ft). Perimeter constraint: $2x + 2y \leq 100$.

Step2: Simplify and solve for length

$x + y \leq 50 \implies y \leq 50 - x$

Step3: Write area function

Area $A(x) = x \cdot y = x(50 - x) = -x^2 + 50x$ (for maximum area)

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Problem 4

Step1: Define variables

Let $x$ = width of the quilt (ft), $y$ = length of the quilt (ft). Piping covers 2 widths + 1 length: $2x + y = 16$.

Step2: Solve for length

$y = 16 - 2x$

Step3: Write area function

Area $A(x) = x \cdot y = x(16 - 2x) = -2x^2 + 16x$

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Problem 5

Step1: Define variables

Let $x$ = width of the garden (m), $y$ = length of the garden (m). Fencing covers 2 widths + 1 length (home is the 4th side): $2x + y = 24$.

Step2: Solve for length

$y = 24 - 2x$

Step3: Write area function

Area $A(x) = x \cdot y = x(24 - 2x) = -2x^2 + 24x$

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Problem 6

Step1: Define variables

Let $x$ = width of the rink (m), $y$ = length of the rink (m). Perimeter constraint: $2x + 2y \leq 250$.

Step2: Simplify and solve for length

$x + y \leq 125 \implies y \leq 125 - x$

Step3: Write area function

Area $A(x) = x \cdot y = x(125 - x) = -x^2 + 125x$ (for maximum area)

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Answer:

1. $A(x) = -2x^2 + 300x$ (where $x$ = width of the parking lot in feet)
2. $A(x) = -x^2 + 20x$ (where $x$ = width of the garden in meters)
3. $A(x) = -x^2 + 50x$ (where $x$ = width of the sandbox in feet, for maximum area)
4. $A(x) = -2x^2 + 16x$ (where $x$ = width of the quilt in feet)
5. $A(x) = -2x^2 + 24x$ (where $x$ = width of the garden in meters)
6. $A(x) = -x^2 + 125x$ (where $x$ = width of the ice rink in meters, for maximum area)