Math Problem Statement

You have 700 feet of fencing to enclose a rectangular plot that borders on a river. If you do not fence the side along the​ river, find the length and width of the plot that will maximize the area. What is the largest area that can be​ enclosed? 700 minus 2 x River x x Question content area bottom Part 1 The​ width, labeled x in the​ figure, is    enter your response here feet.  

Solution

This problem involves maximizing the area of a rectangle, given a constraint on the amount of fencing available. You have 700 feet of fencing to enclose three sides of a rectangular plot (two widths and one length), with one side (the length) along the river unfenced.

Step-by-step solution:

  1. Define Variables: Let:

    • xx be the width of the rectangle (the sides perpendicular to the river).
    • The length of the rectangle, yy, is the side parallel to the river (this side is unfenced).
  2. Constraint Equation: You have 700 feet of fencing for the two widths and one length. Therefore, the total fencing is: 2x+y=7002x + y = 700 Solving for yy: y=7002xy = 700 - 2x

  3. Area Equation: The area AA of the rectangle is given by the product of the width xx and the length yy: A=xyA = x \cdot y Substituting y=7002xy = 700 - 2x into the area equation: A(x)=x(7002x)A(x) = x \cdot (700 - 2x) Simplifying: A(x)=700x2x2A(x) = 700x - 2x^2 This is the quadratic equation for the area.

  4. Maximizing the Area: To maximize the area, we need to find the value of xx that gives the maximum value of A(x)A(x). This is done by finding the vertex of the parabola described by A(x)A(x). For a quadratic equation A(x)=ax2+bx+cA(x) = ax^2 + bx + c, the maximum or minimum occurs at: x=b2ax = -\frac{b}{2a} Here, a=2a = -2 and b=700b = 700, so: x=7002(2)=7004=175x = -\frac{700}{2(-2)} = \frac{700}{4} = 175 Thus, the width xx that maximizes the area is 175175 feet.

  5. Finding the Length: Now that we know x=175x = 175, substitute this value into the equation for yy: y=7002(175)=700350=350 feety = 700 - 2(175) = 700 - 350 = 350 \text{ feet}

  6. Maximized Area: The largest area that can be enclosed is: A=175350=61,250 square feetA = 175 \cdot 350 = 61,250 \text{ square feet}

Final Answer:

  • Width: 175175 feet
  • Length: 350350 feet
  • Largest Enclosed Area: 61,25061,250 square feet

Would you like more details or clarifications on any step?

Here are 5 related questions to consider:

  1. How would the solution change if all four sides were to be fenced?
  2. What would the maximum area be if you had 1,000 feet of fencing instead of 700 feet?
  3. How does the derivative of the area function help determine the maximum?
  4. How can you verify that the solution gives a maximum area and not a minimum?
  5. What happens to the enclosed area if the river side also needs fencing?

Tip: Always check whether a quadratic function opens upward or downward by examining the sign of the coefficient of x2x^2. If it’s negative, the function opens downward, indicating a maximum.

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Math Problem Analysis

Mathematical Concepts

Optimization
Quadratic Equations
Geometry

Formulas

Area of rectangle A = x * y
Fencing constraint: 2x + y = 700
Maximizing quadratic function: A(x) = 700x - 2x^2
Vertex of a parabola: x = -b/(2a)

Theorems

Quadratic Optimization Theorem: Maximum or minimum of a quadratic function occurs at the vertex
Basic Area Theorem for rectangles

Suitable Grade Level

Grades 10-12