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This series of problems will build your fluency with writing and solving quadratic functions to model contextual scenarios. For each question below, write a quadratic function to model the scenario, and then use the graph of the function to answer the questions. ● Use the answer key to determine that you accurately wrote the functions. ● If necessary, revise your work. 1. Farmer Pam plans to fence a garden in her backyard. She buys 80 feet of fencing and plans to enclose a rectangular space where her existing backyard fence forms one side of the enclosure. Farmer Pam wants to maximize the area enclosed by the fence. What is the maximum area she can enclose? 2. In Olympic Women's High Diving, women dive from a platform 20 meters high. The acceleration from gravity is – 9.8 meters per second squared, and their initial velocity is 3 meters per second. If the equation for projectile motion is 푓(푡) = 1 2 푎푡 2 +푣0 푡+ℎ, how long do the divers take to reach the water's surface?

© Success Academy Charter Schools 2 3. The length of a rectangle is two more than its width. The area of the rectangle is 63 cm 2 . Find the dimensions of the rectangle. 4. Best Pools Ever is building a new community pool in Harlem. The pool is surrounded by a 2-foot wide walkway. The width of the pool is 23 feet less than the length of the pool. The total area enclosed by the outer edge of the walkway is 1674 ft 2 . What are the dimensions of the pool?

1 answer

Let's solve each scenario step by step, writing the quadratic functions where necessary and finding the required dimensions or maximum areas.

Problem 1:

Farmer Pam is fencing a rectangular garden with one side against an existing fence.

  1. Define Variables:

    • Let \( x \) be the length of the two sides perpendicular to the existing fence.
    • The width (the side parallel to the existing fence) will then be \( 80 - 2x \) because she has 80 feet of fencing.

  2. Area Function:

    • The area \( A \) is given by: \[ A(x) = x(80 - 2x) = 80x - 2x^2 \]

  3. Find Maximum Area:

    • This is a downward-opening parabola (the coefficient of \( x^2 \) is negative), so the maximum area occurs at the vertex. The vertex \( x \) of a parabola \( ax^2 + bx + c \) is found using: \[ x = -\frac{b}{2a} = -\frac{80}{2(-2)} = 20 \]
    • Substitute \( x = 20 \) back into the area function: \[ A(20) = 20(80 - 2(20)) = 20(40) = 800 \]
    • Maximum Area: 800 square feet.

Problem 2:

For the Olympic Women's High Diving.

  1. Equation of Motion:

    • The equation is given as: \[ f(t) = \frac{1}{2}(-9.8)t^2 + 3t + 20 \]

  2. Find Time to Reach Water's Surface:

    • Set \( f(t) = 0 \): \[ 0 = -4.9t^2 + 3t + 20 \]
    • Rearranging gives: \[ 4.9t^2 - 3t - 20 = 0 \]

  3. Use Quadratic Formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(4.9)(-20)}}{2(4.9)} \]

    • Calculate the discriminant: \[ b^2 - 4ac = 9 + 392 = 401 \]
    • Finding \( t \): \[ t = \frac{3 \pm \sqrt{401}}{9.8} \]
    • Calculating gives two values, consider only the positive value (time must be positive): \[ t \approx 2.26 \text{ seconds} \]

Problem 3:

Finding dimensions of a rectangle where length is two more than width.

  1. Variables:

    • Let \( w \) be the width and \( l = w + 2 \) be the length.

  2. Area Equation:

    • The area is given by: \[ A = lw = w(w + 2) = 63 \]
    • Expanding gives: \[ w^2 + 2w - 63 = 0 \]

  3. Find Roots Using Quadratic Formula: \[ w = \frac{-2 \pm \sqrt{2^2 - 4(1)(-63)}}{2(1)} = \frac{-2 \pm \sqrt{256}}{2} = \frac{-2 \pm 16}{2} \]

    • The possible \( w \) values: \[ w = 7 \quad (\text{positive}), \quad w = -9 \quad (\text{not valid}) \]
    • Dimensions are width \( 7 \) cm and length \( 9 \) cm.

Problem 4:

For Best Pools Ever.

  1. Variables:

    • Let \( l \) be the length, then the width \( w = l - 23 \).

  2. Total Area:

    • The area including the walkway: \[ (l + 4)(w + 4) = 1674 \implies (l + 4)((l - 23) + 4) = 1674 \]
    • Becomes: \[ (l + 4)(l - 19) = 1674 \]
    • Expanding leads to: \[ l^2 - 15l - 1674 = 0 \]

  3. Solve using Quadratic Formula: \[ l = \frac{15 \pm \sqrt{(-15)^2 - 4(1)(-1674)}}{2(1)} \]

    • Calculate: \[ l = \frac{15 \pm \sqrt{2529}}{2} \]

  4. Find Dimensions:

    • Dimensions \( l \) and calculate \( w \) once \( l \) value is found.

You can substitute these solutions into each scenario to confirm they fit the problem stipulations if needed. Let me know if you'd like further clarifications!