A company plans to enclose three parallel rectangular areas for sorting returned goods. The three areas are within one large rectangular area and 952 yd of fencing is available. What is the largest total area that can be enclosed? A rectangle has 2 horizontal sides and 2 vertical sides. Two vertical line segments divide the rectangle into 3 smaller rectangles.

A(max)  = 28322 yd²

Dimensions:

x  =  238 yd

y  = 119 yd

Step-by-step explanation:

Let  call "x" and "y" horizontal and vertical sides of the rectangle,  then we have:

The total area, sum of areas of the three small rectangles is:

A(r) = x*y

And the length to be fenced is

P = 2*x  +  2*y  * 2*y

P = 2*x  + 4*y        and      952  = 2*x + 4*y     ⇒  y  = ( 952 - 2*x) / 4

Total area as function of x s:

A(r)  =  x * y     ⇒  A(x) =  x* ( 952 - 2*x) / 4

A(x)  =  238*x - (1/2)*x²

Taking derivatives on both sides of the equation we get:

A´(x)  = 238  - x     ⇒   A´(x) = 0      ⇒   238  - x = 0

x = 238 yd

Therefore  y =  ( 952 - 2*x) / 4

y  = ( 952  - 2* 238 ) / 4

y  =  119 yd

Largest area is:

A(max)  =  y * x  

A(max)  = 238 * 119

A(max)  = 28322 yd²

answer:   the paragraph proof is mentioned below.

step-by-step explanation:

given: in a δxwy where

such that, ∠wzx ≅ ∠yzx and zw≅zy

we have to prove that: segment zx is a perpendicular bisector of segment wy.

prove: since, zw≅zy ⇒ ∠wzy is a supplementary angle.

then, ∠wzy= 180°

⇒∠wzx + ∠yzx = 180° ( because xz divides ∠wzy into two parts)

but, ∠wzx ≅ ∠yzx ( given)

⇒∠wzy+∠wzy=180° ( by substitution)

⇒2∠wzy=180°

⇒∠wzy=90°

⇒∠wzy=∠yzx =90°

therefore, xz⊥wy where wz=zy

⇒zx is a perpendicular bisector of wy.