We are tasked with finding the dimensions of a rectangular garden that will maximize its area, given that Emma has 150 meters of fencing material and must leave a 10-meter opening for the entrance on one side.

Let’s define:

• $x$ = the length of the garden (the side that includes the 10-meter entrance).
• $y$ = the width of the garden.

Total perimeter condition:

Since there is a 10-meter opening on one side, the remaining fencing for that side is $x - 10$. The total fencing for the other three sides is: $x + 2y = 150 - 10 = 140$ So the perimeter equation becomes: $x + 2y = 140$ This simplifies to: $x = 140 - 2y$

Area function:

The area $A$ of the garden is: $A = x \times y$ Substituting $x = 140 - 2y$ into the area equation: $A(y) = (140 - 2y) \times y = 140y - 2y^2$ We now have a quadratic equation for the area in terms of $y$: $A(y) = -2y^2 + 140y$

Maximizing the area:

To find the value of $y$ that maximizes the area, we take the derivative of $A(y)$ and set it equal to zero: $\frac{dA}{dy} = -4y + 140$ Setting this equal to zero: $-4y + 140 = 0$ $y = 35$

Finding the corresponding length:

Now that we know $y = 35$, we substitute this into the perimeter equation to find $x$: $x = 140 - 2(35) = 140 - 70 = 70$

Conclusion:

The dimensions of the garden that maximize the area are:

• Length = 70 meters
• Width = 35 meters

Would you like further clarification on the process or details?

Here are 5 related questions:

1. How would the maximum area change if the total fencing was increased?
2. What would happen to the area if the entrance size was changed?
3. How would the dimensions change if only a portion of the perimeter was fenced?
4. Can you generalize this method for a rectangular enclosure with an opening on any side?
5. How could you solve this if the garden had a different shape, such as a triangle?

Tip: Always remember that maximizing the area of a rectangle often involves solving a quadratic equation when constraints are involved.