Math Problem Statement
A farmer wants to fence a rectangular field along a river. He needs no fence along the riverside and has 600 meters of fencing available. What dimensions should he use to maximize the area of the field?
Solution
To maximize the area of a rectangular field with one side along a river (where no fencing is needed on that side), we need to determine the optimal dimensions given a limited amount of fencing material.
Let:
- represent the width of the field perpendicular to the river.
- represent the length of the field parallel to the river.
Since fencing is required for only three sides (two widths and one length), the fencing constraint can be set up as follows:
Our goal is to maximize the area of the field, which can be represented as:
Step 1: Express in terms of using the fencing constraint.
From , solve for :
Step 2: Substitute into the area formula.
Substitute into :
Step 3: Maximize the area function.
To find the maximum area, take the derivative of with respect to and set it to zero: Solving for :
Step 4: Find using .
Substitute back into the equation :
Conclusion
To maximize the area, the farmer should use the dimensions:
- Width meters
- Length meters
Maximum Area
The maximum area achievable with these dimensions is:
Would you like further details or have any questions?
Here are five related questions to explore further:
- How would the solution change if fencing was needed along all four sides?
- What if the available fencing was different? How would that affect the maximum area?
- How would the dimensions change if the field was triangular?
- What is the effect of doubling the available fencing material?
- How would the solution be altered if the field had to be a square?
Tip: Remember that in optimization problems, using constraints to express variables in terms of one another simplifies finding maximum or minimum values.
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Math Problem Analysis
Mathematical Concepts
Optimization
Quadratic Functions
Formulas
Perimeter constraint for three sides: 2x + y = 600
Area of a rectangle: A = x * y
Expression for y in terms of x: y = 600 - 2x
Quadratic form of the area function: A = 600x - 2x^2
Theorems
Optimization principle for quadratic functions
Suitable Grade Level
Grades 10-12
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