Math Problem Statement

Ivana has 240 meters of fencing and wishes to form three sides of a rectangular field. The fourth side borders a river and will not need fencing. One of the sides has length x (in meters).

a) Find a function that gives the area A(x) of the field (in square meters) in terms of x.

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

Let's solve this step by step.

Problem Recap:

  • Ivana has 240 meters of fencing.
  • She wants to form three sides of a rectangular field, with one of the sides (the fourth) bordering a river (which does not need fencing).
  • The length of one of the sides is xx meters.

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

Let:

  • xx represent the length of one side parallel to the river.
  • The total fencing available is 240 meters, which will be used for the other two sides (both of length xx) and one side perpendicular to the river (width).

The fencing used for the two lengths and one width is: 2x+y=2402x + y = 240 where yy is the width of the field (perpendicular to the river).

Solve for yy: y=2402xy = 240 - 2x

The area A(x)A(x) of the rectangle is the product of the length xx and the width yy: A(x)=x×y=x(2402x)=240x2x2A(x) = x \times y = x(240 - 2x) = 240x - 2x^2

Thus, the area function is: A(x)=240x2x2A(x) = 240x - 2x^2


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

To find the value of xx that maximizes the area, we take the derivative of the area function and set it equal to zero (to find the critical points):

dAdx=2404x\frac{dA}{dx} = 240 - 4x

Set the derivative equal to zero: 2404x=0240 - 4x = 0 4x=2404x = 240 x=60x = 60

Thus, the value of xx that maximizes the area is 60 meters.


Part (c): Find the maximum area

Now, substitute x=60x = 60 into the area function A(x)=240x2x2A(x) = 240x - 2x^2:

A(60)=240(60)2(60)2A(60) = 240(60) - 2(60)^2 A(60)=144002(3600)A(60) = 14400 - 2(3600) A(60)=144007200A(60) = 14400 - 7200 A(60)=7200 square metersA(60) = 7200 \text{ square meters}

Thus, the maximum area that the field can have is 7200 square meters.


Summary:

  • (a) The area function is A(x)=240x2x2A(x) = 240x - 2x^2.
  • (b) The side length x=60x = 60 meters gives the maximum area.
  • (c) The maximum area is 7200 square meters.

Would you like more details on any part of this solution, or have any questions? Here are five related questions that could extend this concept:

  1. What would happen to the area if the total fencing were less than 240 meters?
  2. How would the area function change if all four sides needed to be fenced?
  3. What happens to the maximum area if Ivana had 300 meters of fencing instead?
  4. How do we know that the value of xx we found gives a maximum, not a minimum?
  5. What would be the area if xx were 30 meters instead of 60?

Tip: The maximum or minimum of a quadratic function ax2+bx+cax^2 + bx + c occurs at x=b2ax = -\frac{b}{2a}. This is useful for solving similar problems quickly!

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Functions
Optimization

Formulas

Area function: A(x) = 240x - 2x^2
Derivative: dA/dx = 240 - 4x
Maximum/minimum of quadratic function: x = -b/2a

Theorems

Quadratic Optimization
First Derivative Test

Suitable Grade Level

Grades 9-11