Math Problem Statement

A set of fenced-in areas, as shown in the diagram, is being planned on an open field. A total of 900 m of fencing is available. What values of x and y will maximize the overall area of the entire closure? use the derivative to calculate this question within 5 steps

Solution

To solve this optimization problem using derivatives, let's first define the total area and the perimeter in terms of the given constraints and variables xx and yy. We're aiming to maximize the total area subject to a constraint on the amount of fencing available.

Step 1: Define the relationships.

Let xx and yy be the dimensions of the fenced areas.

  1. The total fencing available is 900 meters.
  2. Suppose the diagram suggests a rectangular shape subdivided into smaller areas (commonly into 3 areas for such problems). Assume the structure has:
    • Two outer boundaries of length yy,
    • Three inner boundaries of length xx, as each xx would have its own division.

Thus, the total fencing constraint (perimeter) is given by: 2y+3x=9002y + 3x = 900

Step 2: Define the objective function (area).

The area AA of the entire closure can be expressed as: A=x×yA = x \times y We need to maximize this area subject to the fencing constraint.

Step 3: Solve the constraint for one variable.

From the constraint equation 2y+3x=9002y + 3x = 900, solve for yy in terms of xx: y=9003x2y = \frac{900 - 3x}{2}

Step 4: Substitute into the area formula.

Substitute y=9003x2y = \frac{900 - 3x}{2} into the area equation: A(x)=x×9003x2=900x3x22A(x) = x \times \frac{900 - 3x}{2} = \frac{900x - 3x^2}{2} Simplifying: A(x)=450x3x22A(x) = 450x - \frac{3x^2}{2}

Step 5: Differentiate and find the critical points.

Differentiate the area function with respect to xx: dAdx=4503x\frac{dA}{dx} = 450 - 3x Set dAdx=0\frac{dA}{dx} = 0 to find the critical points: 4503x=0450 - 3x = 0 Solving for xx: x=150x = 150

Now, substitute x=150x = 150 back into the equation for yy: y=9003(150)2=9004502=225y = \frac{900 - 3(150)}{2} = \frac{900 - 450}{2} = 225

Thus, the values of x=150x = 150 meters and y=225y = 225 meters will maximize the area of the closure.

Summary:

The dimensions that maximize the area are: x=150 meters,y=225 meters.x = 150 \text{ meters}, \quad y = 225 \text{ meters}.

Let me know if you'd like more details or have any questions.

5 Related Questions:

  1. How do we verify if this critical point is a maximum or minimum?
  2. What happens to the area if more fencing is added?
  3. Can we solve this optimization problem using Lagrange multipliers?
  4. What would the shape look like if we didn't have the inner divisions?
  5. How would the solution change if the shape were circular instead of rectangular?

Tip:

Always check the second derivative to confirm if a critical point corresponds to a maximum or minimum.

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Math Problem Analysis

Mathematical Concepts

Optimization
Differentiation
Algebra

Formulas

Total fencing constraint: 2y + 3x = 900
Area of the enclosure: A = x * y
Area in terms of x: A(x) = 450x - (3x^2)/2
Derivative: dA/dx = 450 - 3x

Theorems

Maxima and Minima using Derivatives

Suitable Grade Level

Grade 10-12