Question

A company plans to enclose three parallel rectangular areas for sorting returned goods. The three areas are within one large rectangular area and 1088 yd of fencing is available. What is the largest total area that can be​ enclosed?

Answer 1

Answer:

The largest total area that can be enclosed will be a square of length 272 yards.

Step-by-step explanation:

First we get the perimeter of the large rectangular enclosure.

Perimeter of a rectangle =2(l + w)

Perimeter of the large rectangular enclosure= 1088 yard

Therefore:

2(L+W)=1088

The region inside the fence is the area

Area: A = LW

We need to solve the perimeter formula for either the length or width.

2L+ 2W= 1088 yd

2W= 1088– 2L

W = $\frac{1088-2L}{2}$

W = 544–L

Now substitute W = 544–L into the area formula

A = LW

A = L(544 – L)

A = 544L–L²

Since A is a quadratic expression, we re-write the expression with the exponents in descending order.

A = –L²+544L

Next, we look for the value of the x coordinate

$L= -\frac{b}{2a}$

$L= -\frac{544}{2X-1}$

L=272 yards

Plugging L=272 yards into the calculation for area:

A = –L²+544L

A(272)=-272²+544(272)

=73984 square yards

Thus the largest area that could be encompassed would be a square where each side has a length of 272 yards and a width of:

W = 544 – L

= 544 – 272

= 272 yards