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.  

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:
   - \( x \) be the width of the rectangle (the sides perpendicular to the river).
   - The length of the rectangle, \( y \), 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:
   \[
   2x + y = 700
   \]
   Solving for \( y \):
   \[
   y = 700 - 2x
   \]

3. **Area Equation**:
   The area \( A \) of the rectangle is given by the product of the width \( x \) and the length \( y \):
   \[
   A = x \cdot y
   \]
   Substituting \( y = 700 - 2x \) into the area equation:
   \[
   A(x) = x \cdot (700 - 2x)
   \]
   Simplifying:
   \[
   A(x) = 700x - 2x^2
   \]
   This is the quadratic equation for the area.

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

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

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

### Final Answer:
- **Width**: \( 175 \) feet
- **Length**: \( 350 \) feet
- **Largest Enclosed Area**: \( 61,250 \) 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 \(x^2\). If it’s negative, the function opens downward, indicating a maximum.

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?

Learn how to maximize the area of a rectangular plot with 700 feet of fencing, excluding one side along a river. The solution covers optimization techniques, including the use of quadratic equations and geometric concepts. Suitable for high school students in Grades 10-12.

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