Math Problem Statement

Misha has a new rabbit that she named Wascal. She wants to build Wascal a pen so that the rabbit has space to move around safely. Misha has purchased a 72-foot roll of fencing to build a rectangular pen. 1. If Misha uses the whole roll of fencing, what are some of the possible dimensions of the pen? 2. If Misha wants a pen with the largest possible area, what dimensions should she use for the sides? Justify your answer. 3. Write a model for the area of the rectangular pen in terms of the length of one side. Include both an equation and a graph.

Solution

Let's break down the questions based on the image you uploaded and solve them step by step.

  1. If Misha uses the whole roll of fencing, what are some of the possible dimensions of the pen?

    The total perimeter of the rectangular pen is 72 feet. The perimeter of a rectangle is given by the formula:

    P=2l+2wP = 2l + 2w

    Where ll is the length, and ww is the width. Since the total fencing is 72 feet, we have:

    2l+2w=722l + 2w = 72

    Dividing the equation by 2:

    l+w=36l + w = 36

    So, the sum of the length and width must be 36 feet. There are many possible dimensions, such as:

    • If l=30l = 30, then w=6w = 6
    • If l=20l = 20, then w=16w = 16
    • If l=25l = 25, then w=11w = 11

    These are just a few examples. The length and width must always add up to 36 feet.

  2. If Misha wants a pen with the largest possible area, what dimensions should she use for the sides? Justify your answer.

    To maximize the area, we use the area formula for a rectangle:

    A=l×wA = l \times w

    From the equation l+w=36l + w = 36, we can express ww as:

    w=36lw = 36 - l

    Substituting this into the area formula:

    A=l×(36l)A = l \times (36 - l)

    Expanding this:

    A=36ll2A = 36l - l^2

    This is a quadratic equation. To maximize the area, we can find the vertex of this parabola. The vertex form of a quadratic equation Ax2+Bx+CAx^2 + Bx + C has its maximum at x=B2Ax = -\frac{B}{2A}.

    Here, A=1A = -1 and B=36B = 36, so the length that maximizes the area is:

    l=362(1)=18l = -\frac{36}{2(-1)} = 18

    If l=18l = 18, then w=3618=18w = 36 - 18 = 18.

    Therefore, the dimensions that give the largest possible area are 18 feet by 18 feet, which forms a square.

  3. Write a model for the area of the rectangular pen in terms of the length of one side. Include both an equation and a graph.

    The equation for the area in terms of the length ll is:

    A(l)=36ll2A(l) = 36l - l^2

    This is a quadratic equation representing a parabola. The graph of this equation will show the maximum area at l=18l = 18, where the area reaches its peak.

Would you like me to generate the graph for the area function as well?

Questions for Deeper Understanding:

  1. What would happen to the area if one side were extremely small, say 1 foot long?
  2. How does the shape of the pen (square vs. rectangle) affect the area?
  3. What other shapes might maximize the area for a given perimeter besides rectangles?
  4. How does the quadratic function help in solving maximization problems like this one?
  5. Can the same method be applied to different shapes, like circles or triangles, to maximize the area?

Tip: In problems involving maximizing or minimizing areas or other quantities, look for ways to express one variable in terms of another to simplify the calculations.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Optimization
Perimeter and Area

Formulas

Perimeter of a rectangle: P = 2l + 2w
Area of a rectangle: A = l * w
Quadratic formula to find maximum area: A(l) = 36l - l^2

Theorems

Vertex of a parabola: The maximum area occurs at the vertex of a quadratic function

Suitable Grade Level

Grades 8-10