# The management of a large store has 1000 feet of fencing to enclose a rectangular storage yard using the building as one side of the yard . If the fencing is used for the remaining 3 sides, find...

The management of a large store has 1000 feet of fencing to enclose a rectangular storage yard using the building as one side of the yard . If the fencing is used for the remaining 3 sides, find the area of the largest possible yard.

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To support the above answer, using calculus is another way to solve this such that, from the derived equation for area in this question:

`A=1000l-2l^2`

as `A'=0` for the maximum area and

`A= 1000l -2l^2` (as explained above)

`therefore A'=1000- 4l`

`therefore 0=1000-4l`

`therefore 4l=1000`

`l=250`

Substitute:

`therefore w=1000-2(250)`

`w=500`

Substitute:

`therefore area = 250 times 500`

**Ans:**

**Maximum Area=125 000 ft^2**

The total amount of fencing is` 1000` ft, we will take note of the formula for the Perimeter of a rectangle.

We know that the formula for the Perimeter of a rectangle is:

`P = 2L + 2W`

For this problem one side is for a building, so the formula that we will consider is:

`P = L + 2W`

Since we are given with` 1000` ft for fencing the equation will be `1000 = L + 2W` (first equation)

We are maximizing the area of it, so we will consider the formula for the area of a rectangle which is:` A = L * W ` (second equation)

We have two equations, we can use substitution method.

Let us use the first equation and solve for L.

`1000 = L + 2W`

Since we are isolating the L on one side, we need to get rid of the 2W on right side.

The 2W is being added on the right side, so we need to use the opposite operation of addition, which is subtraction. So, we subtract 2W on both sides.

`1000 - 2W = L + 2W - 2W`

`L = 1000 - 2W`

Replace the L on the `A = L*W` by `1000 - 2W` .

`A = (1000 - 2W)*W `

Use Distributive Property: `a(b + c) = ab + ac` .

`A = 1000W - 2W^2`

We know that the graph of this equation is a parabola opening upward, which we can

find the maximum point on the vertex `(h, k)` . Therefore, we can find the maximum area by finding the vertex. The `k` will be the maximum area.

Find the value of` h` by the formula `h = (-b)/(2a)` .

`h = (-1000)/(2*-2) = (-1000)/(-4) = 250`

Finding the maximum area:

`A = 1000(250) - 2(250)^2 = 250000 - 125000 = 125000 `

The maximum area is **125,000 ft^2.**