Question

lesson 1 assignment

use a separate piece of paper for your journal entry.

journal
in your own words, explain the
four steps of the mathematical
modeling process.

• notice and wonder
• organize and mathematize
• predict and analyze
• test and interpret

practice
1 mr. jones wants to fence in a rectangular field for his horse using the 600 feet of fence he
has stored in his barn. he wants to maximize the area of the field in order to give his horse
the most pasture possible. help mr. jones design his field to achieve the maximum area.
(a) complete the table to show the length of the field for each given width.
(b) define the function ( ell(w) ) to represent the length of the field as a function of the width.
explain your reasoning.
(c) define the function ( a(w) ) to represent the area of the field as a function of the width.
explain your reasoning.
(d) determine the maximum area of the field as well as the length and width that will result
in the maximum area. explain your reasoning.
go to livehint.com for help
on the practice questions.
lesson 1 > blame it on the rain
117

Explanation:

Step1: Find the function for length

The perimeter of a rectangle is \(P = 2(l + w)\). Given \(P=600\), we have \(600 = 2(l + w)\). Solving for \(l\), we get \(l=\frac{600 - 2w}{2}=300 - w\). So, \(\ell(w)=300 - w\).

Step2: Find the function for area

The area of a rectangle is \(A=l\times w\). Substituting \(l = 300 - w\) into the area formula, we get \(A(w)=(300 - w)w=-w^{2}+300w\).

Step3: Find the maximum of the quadratic function

For a quadratic function \(y = ax^{2}+bx + c\) (\(a=- 1\), \(b = 300\), \(c = 0\)), the vertex of the parabola (which gives the maximum for \(a<0\)) has \(w=-\frac{b}{2a}\). Substituting \(a=-1\) and \(b = 300\), we have \(w=-\frac{300}{2\times(-1)} = 150\).
When \(w = 150\), \(l=300 - 150=150\). And \(A(150)=-150^{2}+300\times150=-22500 + 45000=22500\).

Answer:

• Function for length: \(\ell(w)=300 - w\). The reasoning is based on the perimeter formula of a rectangle \(P = 2(l + w)\) and solving for \(l\) when \(P = 600\).
• Function for area: \(A(w)=-w^{2}+300w\). Since \(A=l\times w\) and \(l = 300 - w\) (from the perimeter - length relationship), we substitute \(l\) into the area formula.
• Maximum area and dimensions: The maximum area \(A = 22500\) square feet occurs when \(w = 150\) feet and \(l=150\) feet. For a quadratic function \(y = ax^{2}+bx + c\) (\(a=-1\), \(b = 300\)), the \(x\) - coordinate of the vertex \(w=-\frac{b}{2a}\) gives the value of \(w\) that maximizes the area. Substituting \(w = 150\) into the length and area formulas gives the corresponding length and maximum area.