Math Problem Statement

Lisa has 320 meters of fencing available to enclose three sides of a rectangular field. One side of the rectangle borders a river and will not need fencing. (a) Find a function that gives the area A(x) of the field (in square meters). (b) What side length x gives the maximum area that the field can have? (c) What is the maximum area that the field can have?

Solution

The problem in the image involves maximizing the area of a rectangular field with a fixed amount of fencing, 320 meters, and one side along a river (which doesn't need fencing). Let's break down the question step-by-step.

Given Information:

  • The total length of fencing available is 320 meters.
  • One side of the rectangle is along the river, so no fencing is needed on that side.
  • The two other sides are of equal length (let's call it xx), and the opposite side (parallel to the river) is yy.

Steps to Solve:

Part (a): Find a function for the area A(x)A(x)

  • The total perimeter with fencing is made up of 2 sides of length xx and one side of length yy (since the side along the river doesn't need fencing). 2x+y=3202x + y = 320
  • Solve for yy: y=3202xy = 320 - 2x
  • The area A(x)A(x) of the rectangle is: A(x)=xy=x(3202x)A(x) = x \cdot y = x(320 - 2x) Therefore, the area function is: A(x)=320x2x2A(x) = 320x - 2x^2

Part (b): Find the value of xx that maximizes the area

To find the value of xx that maximizes the area, we take the derivative of A(x)A(x) with respect to xx and set it equal to zero.

A(x)=3204xA'(x) = 320 - 4x Set the derivative equal to 0: 3204x=0x=80320 - 4x = 0 \quad \Rightarrow \quad x = 80

So, the value of xx that maximizes the area is x=80x = 80 meters.

Part (c): What is the maximum area?

To find the maximum area, substitute x=80x = 80 into the area function: A(80)=320(80)2(80)2=2560012800=12800square metersA(80) = 320(80) - 2(80)^2 = 25600 - 12800 = 12800 \, \text{square meters}

Final Answer:

  • (a) The area function is A(x)=320x2x2A(x) = 320x - 2x^2.
  • (b) The side length xx that gives the maximum area is 80 meters.
  • (c) The maximum area is 12,800 square meters.

Would you like further clarification on any of the steps, or have additional questions?

Here are 5 related questions to expand on this concept:

  1. How would the problem change if there was fencing on all four sides of the rectangle?
  2. How do we verify that the critical point x=80x = 80 is indeed a maximum?
  3. Can you find the dimensions of the rectangle if the total amount of fencing was 400 meters instead of 320 meters?
  4. What happens to the area if the river doesn't exist, and all four sides must be fenced?
  5. Can you generalize the formula for maximizing the area given any total fencing length?

Tip: For maximization problems, the first derivative test is an efficient way to find local maxima or minima, but always check the second derivative or boundary conditions to confirm if it's truly a maximum.

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Math Problem Analysis

Mathematical Concepts

Optimization
Calculus
Quadratic Functions

Formulas

Perimeter constraint: 2x + y = 320
Area function: A(x) = x * y
Maximization using the derivative: A'(x) = 0

Theorems

First Derivative Test for Maxima and Minima

Suitable Grade Level

Grades 10-12