Question

Ramon wants to fence in a rectangular portion of his back yard against the back of his garage for a vegetable garden. he plans to use 40 feet of fence, and needs fence on only three sides. find the maximum area he can enclose. (hint: the lengths of the 3 fenced sides of the rectangle must add up to 40.)

Answer 1

To find the maximum area Ramon can enclose with 40 feet of fence on three sides, we set up an equation for the area in terms of one variable and used calculus to find that the maximum area is 200 square feet when the length is 20 feet and the width is 10 feet.

Step-by-step explanation:

Ramon wants to fence in a rectangular portion of his back yard for a vegetable garden, using 40 feet of fencing material for three sides. Since the back of his garage forms one side of the rectangle, he only needs to fence the other three sides. To find the maximum area he can enclose, we should use the formula for the perimeter of a rectangle, P = 2l + 2w, where l is length and w is width. However, since one side is provided by the garage, the perimeter formula for this scenario is P = l + 2w. Given P is 40 feet, we have l + 2w = 40.

To maximize the area, A = l * w, we need to find the optimal values of l and w. By turning the problem into a function of a single variable, we can use calculus or other optimization techniques to find the maximum. A simplification, given that we know the perimetric constraint, is to express w in terms of l or vice versa. Let's express the width w as (40-l)/2.

Substitute w into the area function: A(l) = l * ((40 - l)/2). This simplifies to A(l) = 20l - (l^2)/2. To find the maximum area, we take the derivative of A with respect to l and set it to zero:

A'(l) = 20 - l = 0
Solving this gives us l = 20 feet. Then, w = (40 - l) / 2 = (40 - 20) / 2 = 10 feet.

The maximum area he can enclose with 40 feet of fence on three sides is therefore 20 feet * 10 feet = 200 square feet.

Answer 2

The first thing to do is find the perimeter of the fence:
l + 2w = 40,
l = length
w = width
By definition the area of a rectangle is:
A = l * w
Clearing the perimeter found:
l = 40-2w
We substitute l in the expression of the area:
A = w * (40-2w)
We rewrite:
A = -2w ^ 2 + 40w
We observe that it is a parabola that opens downwards
We find the maximum of the function:
A '= - 4w + 40 = 0
4w = 40
w = 10
Substituting in the expression of length:
l = 40-2 (10) = 20
The maximum area is then:
A = (10) * (20) = 200 feet ^ 2
Answer:
the maximum area he can enclose is 200 feet ^ 2