Math Problem Statement

Farmer Ed has 5,500 meters of fencing and wants to enclose a rectangular plot that borders on a river. If Farmer Ed does not fence the side along the river, what is the largest area that can be enclosed?

Solution

Let's solve this problem step-by-step.

Problem Analysis

Farmer Ed wants to enclose a rectangular plot that borders a river, using only three sides of fencing (two widths and one length). He has 5,500 meters of fencing to use, and the side along the river will not be fenced.

Let:

  • xx = the length of the two sides perpendicular to the river.
  • The length along the river is 55002x5500 - 2x.

Step 1: Define the Area Function

The area AA of the rectangular plot can be expressed in terms of xx: A=x×(55002x)A = x \times (5500 - 2x) Simplify the equation: A=5500x2x2A = 5500x - 2x^2

Step 2: Find the Maximum Area

To find the maximum area, we need to express AA as a quadratic function and determine its maximum value.

The function is: A(x)=2x2+5500xA(x) = -2x^2 + 5500x

Step 3: Find the Vertex of the Parabola

The quadratic function A(x)=2x2+5500xA(x) = -2x^2 + 5500x is a downward-opening parabola. The maximum value occurs at the vertex. The formula for the vertex of a parabola ax2+bx+cax^2 + bx + c is: x=b2ax = -\frac{b}{2a} In our case:

  • a=2a = -2
  • b=5500b = 5500

So, x=55002×(2)=55004=1375x = -\frac{5500}{2 \times (-2)} = \frac{5500}{4} = 1375

Step 4: Calculate the Maximum Area

Now substitute x=1375x = 1375 back into the area equation to find the maximum area: A=5500x2x2A = 5500x - 2x^2 A=5500(1375)2(1375)2A = 5500(1375) - 2(1375)^2 A=75625002×1890625A = 7562500 - 2 \times 1890625 A=75625003781250=3781250A = 7562500 - 3781250 = 3781250

Final Answer

The largest area that can be enclosed is 3,781,250\boxed{3,781,250} square meters.

Would you like me to provide more details or solve any other parts of this problem?

Here are 5 related questions you might find interesting:

  1. How would the solution change if Farmer Ed fenced all four sides instead of just three?
  2. What if Farmer Ed had only 4,000 meters of fencing? How would that change the maximum area?
  3. How does the concept of optimization apply to other practical problems in agriculture?
  4. Can we use calculus to verify the maximum area we found for this problem?
  5. What is the effect of changing the length of fencing available on the shape of the optimal rectangle?

Tip: When maximizing or minimizing a function, look for opportunities to use the vertex formula or calculus-based techniques like finding critical points.

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Math Problem Analysis

Mathematical Concepts

Quadratic Optimization
Area of a Rectangle

Formulas

Area of a rectangle: A = length × width
Quadratic formula for vertex: x = -b/(2a)

Theorems

Vertex formula for parabolas

Suitable Grade Level

Grades 10-12