Q9. A rectangular field has a fence along two opposite sides and a brick wall along the other two sides. Fence costs $5 per meter and brick costs $40 per meter.

• a. Express the total cost of the fence and brick in terms of the length and width of the field.

• b. The area of the field will be 230 m2. Express the total number of meters of fence needed in terms of the width w of the field.

• c. Use this to express the total cost of brick and fence in terms of the width w.

• d. What should the width of the field be so the cost is smallest? (You can leave a square root in your answer.)

Background

Topic: Optimization and Cost Functions in Calculus

This problem tests your ability to set up and optimize a cost function using calculus, given constraints on area and material costs. It involves expressing costs in terms of variables, applying area constraints, and finding minimum values using derivatives.

Key Terms and Formulas

• Length (L): The longer side of the rectangle.

• Width (w): The shorter side of the rectangle.

• Area constraint:

• Cost function: Total cost = (cost per meter of fence) × (total meters of fence) + (cost per meter of brick) × (total meters of brick)

• Optimization: Use calculus to find the minimum of the cost function.

Step-by-Step Guidance

1. Identify which sides of the rectangle are fenced and which are brick. According to the diagram, the two sides of length are brick, and the two sides of width are fence.

2. Write the total cost function in terms of and :

3. Apply the area constraint: . Solve for in terms of :

4. Substitute from the area constraint into the cost function to express total cost in terms of only:

5. Simplify the cost function and set up the derivative with respect to to find the minimum cost. Do not solve for the minimum yet.

Try solving on your own before revealing the answer!