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sdas [7]
3 years ago
7

SOMEONE HELP ASAP i’ll give brainliest

Mathematics
1 answer:
densk [106]3 years ago
4 0

Answer:

which question is it ??

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A company plans to enclose three parallel rectangular areas for sorting returned goods. The three areas are within one large rec
Svetlanka [38]

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

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