Calculus I Practice Test 3 – Step-by-Step Study Guidance

Q3. A dairy farmer plans to make two identical and adjacent rectangular pastures next to a barn. The total area must be 15,000 square meters. What dimensions will require the least amount of fencing?

Background

Topic: Optimization (Applied Maximum/Minimum Problems)

This problem is a classic calculus optimization scenario. You are asked to minimize the amount of fencing (perimeter) needed for a fixed area, using calculus to find the optimal dimensions.

Key Terms and Formulas

• Area constraint: The total area of the two pastures must be 15,000 m2.

• Perimeter (fencing) to minimize: The total length of fencing used for the sides and dividers (not including the barn side).

• Variables: Let x = width of each pasture, y = length parallel to the barn.

• Area formula: $A = 2xy$

• Fencing formula: $F = 3x + 2y$ (since there are three vertical sections and two horizontal sections to fence)

Step-by-Step Guidance

1. Write the area constraint using the variables: $2xy = 15,000$.

2. Solve the area constraint for one variable in terms of the other. For example, solve for $y$ in terms of $x$:

   $y = \frac{15,000}{2x}$

3. Write the fencing (perimeter) function $F$ in terms of a single variable by substituting the expression for $y$ into the fencing formula:

   $F(x) = 3x + 2y = 3x + 2\left(\frac{15,000}{2x}\right)$

4. Simplify the fencing function:

   $F(x) = 3x + \frac{15,000}{x}$

5. To find the minimum fencing, take the derivative of $F(x)$ with respect to $x$, set it equal to zero, and solve for $x$:

   $F'(x) = 3 - \frac{15,000}{x^2}$

Try solving on your own before revealing the answer!

Final Answer:

The minimum fencing occurs when $x = 70.71$ meters and $y = 106.07$ meters (rounded to two decimal places).

This is found by solving $3 - \frac{15,000}{x^2} = 0$ for $x$, then substituting back to find $y$.