Math Problem Statement

A gardener has 400 feet of fencing to fence in a rectangular garden. One side of the garden is bordered by a river and so it does not need any fencing. What dimensions would guarantee that the garden has the greatest possible area?

Solution

This problem involves finding the dimensions of a rectangular garden that will maximize the area, given a fixed amount of fencing. The garden is bordered by a river on one side, so fencing is only needed for three sides.

Step-by-Step Approach:

Let:

  • xx be the length of the side parallel to the river (no fence needed on this side).
  • yy be the length of the other two sides (perpendicular to the river), which need fencing.

The total length of the fence available is 400 feet, and fencing is required only on three sides. Therefore, the equation for the perimeter is: x+2y=400x + 2y = 400 This simplifies to: x=4002yx = 400 - 2y

The area AA of the garden is given by: A=x×yA = x \times y Substitute x=4002yx = 400 - 2y into the area equation: A=(4002y)×y=400y2y2A = (400 - 2y) \times y = 400y - 2y^2

Maximize the Area:

To find the maximum area, take the derivative of AA with respect to yy, and set it equal to zero: dAdy=4004y\frac{dA}{dy} = 400 - 4y Set dAdy=0\frac{dA}{dy} = 0: 4004y=0400 - 4y = 0 Solving for yy: y=100 feety = 100 \text{ feet}

Now, substitute y=100y = 100 into the equation for xx: x=4002(100)=200 feetx = 400 - 2(100) = 200 \text{ feet}

Conclusion:

  • Shorter side (perpendicular to the river): y=100y = 100 feet
  • Longer side (parallel to the river): x=200x = 200 feet
  • Greatest possible area: A=200×100=20,000 square feetA = 200 \times 100 = 20,000 \text{ square feet}

Would you like further details or have any questions?

Here are 5 related questions for deeper understanding:

  1. How would the solution change if all four sides of the garden needed fencing?
  2. Can we derive a general formula for maximizing area with different amounts of available fencing?
  3. How would the problem change if the river were on one of the shorter sides instead of the longer side?
  4. What other mathematical methods can be used to solve this type of optimization problem?
  5. What are the real-world applications of this type of problem?

Tip: Always check boundary conditions in optimization problems to ensure you find the maximum or minimum value.

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Math Problem Analysis

Mathematical Concepts

Optimization
Algebra
Quadratic Functions

Formulas

Perimeter equation: x + 2y = 400
Area equation: A = x * y
Derivative of area: dA/dy = 400 - 4y

Theorems

Optimization using derivatives

Suitable Grade Level

Grades 9-12