Question

A developer wants to enclose a rectangular grassy lot that borders a city street for parking. if the developer has 256 feet of fencing and does not fence the side along the​ street, what is the largest area that can be​ enclosed?

Answer 1

The largest area that can be enclosed is 8192 square feet.

Step-by-step explanation:

To find the largest area that can be enclosed, we need to consider that the developer has 256 feet of fencing and does not fence the side along the street. Let's assume that the length of the side along the street is x feet.

Since there are 3 sides that are fenced, their total length must be 256 - x. Let's divide this length into two equal sides of length (256 - x)/2 each.

The area of a rectangle is given by length * width. In this case, the length is (256 - x)/2 and the width is x. So the area A(x) = x * (256 - x)/2.

Now, let's find the maximum value of A(x) by finding the critical points. To do this, we take the derivative of A(x), which is (256 - 2x)/2, and set it equal to zero. Solving this equation, we find x = 128.

Therefore, the largest area that can be enclosed is A(128) = 128 * (256 - 128)/2 = 8192 square feet.

Answer 2

Let the lenght perpendicular to street = x

Means lenght two sides will be x and x

Another side of retangle will be = 256 - 2x

Area of rectangle will be= A = x * (256-2x) = 256 x - 2 x^2

For maximum area, dA/dx = 0

d/dx (256 x - 2 x^2) = 0

256 - 4 x = 0

4 x = 256

x = 64 feet

Means one side is 64 feet

Another side will be = 256 - 2 * 64 = 128

Now we have two sides of rectangle 64 and 128

Area = 128 * 64 = 8192 square feet : Answer

Hope it will help :)