Acompany plans to enclose three parallel rectangular areas for sorting returned goods. the three areas are within one large rectangular area and 10641064 yd of fencing is available. what is the largest total area that can be enclosed?

First get the perimeter

Perimeter: 2l + 2w = 1064 yd

The region inside the fence is the area

Area: A = lw

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

2l + 2w = 1064 yd

2w = 1064 yd – 2l

W = 1064– 2l / 2

W = 532 – l

Now substitute than to the area formula

A = lw

A = l (532 – l)

A = 532 – l^2

Since the equation A represents a quadratic expression, rewritte the expression with the exponents in descending order

A(l) = -l^2 + 532l

Then look for the value of the x coordinate

l = -b/2a

l = -532/2(-1)

l = -532/-2

l = 266 yards

Plugging in the value into our calculation for area:

A(l) = -l^2 + 532

A(266) = -(266)^2 + 532 (266)

A(266) =  70756+ 141512

= 70756 square yards.

Thus the largest area that could encompass would be a square where each side has a length of 266 yards and a width of:

W = 532 – l

= 532 – 266

= 266

It should be noted that the largest total area that can be enclosed will be 70756 yards².

Firstly, we have to calculate the perimeter. This will be:

2l + 2w = 1064 yd

Area: A = length × width

2l + 2w = 1064 yd

2w = 1064 yd – 2l

W = 1064– 2l / 2

W = 532 – l

Since Area = lw

A = l(532 – l)

A = 532 – l²

A(l) = -l² + 532l

l = -b/2a

l = -532/2(-1)

l = -532/-2

l = 266 yards

The length is 266 yards.

Area = -l² + 532

A( = -(266)² + 532 (266)

Area = 70756+ 141512

Area = 70756 yards²

Learn more about areas on:

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no, the equation would be invalid since they are both not equal.

step-by-step explanation:

here you got

4y-6+4y 2-9

first, you would simplify the equation on the form.

4y-6+4y 2-9

then, move all the terms that contains a y to the left and all the other terms to the right.

+4y-4y+-9-6

finally, you will simplify the left and the right side of the equation.

y=-3