Final answer:
To find the largest possible area of a rectangular garden with one side against the house and 50 ft of fence available, we set up an optimization problem. We then derive the area function with respect to one variable to find the dimensions that yield the maximum area, resulting in an area of 312.5 square feet with dimensions of 12.5 ft by 25 ft.
Step-by-step explanation:
The question is asking for the largest possible area of a rectangular vegetable garden that can be fenced using 50 ft of fence, with one side along the wall of a house. This means we only need to fence off three sides of the rectangle, using the house as the fourth side.
To find the maximum area, we can use the problem-solving strategy that involves optimizing a function of two variables. Since we only need to fence three sides (let's call the lengths of the two shorter sides x and the length of the side parallel to the house y), the perimeter that we can use is 50 ft, which leads us to the equation x + x + y = 50 ft or 2x + y = 50 ft.
To maximize the area A = xy, we need to express y in terms of x using the perimeter equation. So y = 50 - 2x. We can now write the area as a function of x alone: A(x) = x(50 - 2x).
Now, to find the maximum area, we can calculate the derivative of A(x) with respect to x and find where it equals zero (i.e., where the slope of the area function is zero). Doing this, we find that x = 12.5 ft gives the maximum area. Substituting x = 12.5 back into the perimeter equation, we subsequently find that y = 25 ft for the side along the house. Thus, the maximum area that can be fenced is 312.5 square feet.
Note that this result is based on the principle that a rectangle with a given perimeter has the largest area when it is a square. However, since one side is already provided by the house, we deal with a rectangle where one side length (the one along the house) is potentially longer than the other.