Maximizing Area of a Rectangular Enclosure (Optimization Problem)

Q13. The Park and recreational department are planning to build a picnic area for motorists along a quiet dirt road. It has 2000 yards of material to enclose a rectangular area with no fence along the road. What are the dimensions that maximize the area?

Background

Topic: Optimization using Calculus

This question tests your ability to set up and solve an optimization problem involving a rectangular enclosure with one side open (no fence along the road). You need to maximize the area given a fixed amount of fencing material.

Key Terms and Formulas

• Area of rectangle: $A = x y$

• Constraint (fencing): The total fencing used is $2x + y = 2000$ (since the side along the road is open).

• Optimization: Use calculus to maximize the area subject to the constraint.

Step-by-Step Guidance

1. Write the area formula for the rectangle: $A = x y$.

2. Write the constraint equation for the fencing: $2x + y = 2000$.

3. Solve the constraint equation for $y$ in terms of $x$: $y = 2000 - 2x$.

4. Substitute $y$ into the area formula to get $A$ as a function of $x$: $A(x) = x(2000 - 2x)$.

5. Expand and simplify the area function: $A(x) = 2000x - 2x^2$.

Try solving on your own before revealing the answer!

Final Answer: x = 500 yards, y = 1000 yards

To maximize the area, take the derivative of $A(x)$, set it to zero, and solve for $x$. Substitute back to find $y$.

The optimal dimensions are $x = 500$ yards and $y = 1000$ yards.