To maximize the area enclosed by 350 feet of fencing for two adjacent horse corrals, we can set up the problem as follows:

1. Define the variables:

   • Let $x$ be the length of each corral.
   • Let $y$ be the width of the corrals.

2. Write the equation for the perimeter: The total perimeter for two adjacent corrals can be expressed as: $3x + 2y = 350$

3. Express one variable in terms of the other: Rearranging the perimeter equation for $y$: $y = \frac{350 - 3x}{2}$

4. Write the area function: The total area $A$ of the two corrals is: $A = 2xy = 2x \left(\frac{350 - 3x}{2}\right) = x(350 - 3x) = 350x - 3x^2$

5. Find the maximum area: To find the maximum area, take the derivative of $A$ and set it to zero: $\frac{dA}{dx} = 350 - 6x$ Setting the derivative equal to zero: $350 - 6x = 0 \implies 6x = 350 \implies x = \frac{350}{6} \approx 58.33 \text{ feet}$

6. Calculate $y$: Substitute $x$ back into the equation for $y$: $y = \frac{350 - 3 \times 58.33}{2} \approx \frac{350 - 175}{2} = \frac{175}{2} \approx 87.5 \text{ feet}$

7. Conclusion: The dimensions that maximize the area are approximately:

   • Length of each corral $x \approx 58.33$ feet
   • Width of each corral $y \approx 87.5$ feet

Would you like further details on any part of the solution or have any questions? Here are some related questions you might find interesting:

1. How does changing the total length of fencing affect the dimensions for maximum area?
2. What would be the area enclosed with these dimensions?
3. How would you approach this problem if only one corral was required?
4. Can you derive the dimensions using a different method, such as completing the square?
5. How does this scenario relate to the concept of optimization in calculus?

Tip: When dealing with optimization problems, sketching the scenario can often help visualize constraints and solutions.