A company plans to enclose three parallel rectangular areas for sorting returned
ID: 2888592 • Letter: A
Question
A company plans to enclose three parallel rectangular areas for sorting returned goods. The three areas are within one large rectangular area and 1080 yd of fencing is available. What is the largest total area that can be enclosed?
A or b are the options:
a) The length of each side of the larger rectangular area is the same and measures__ yards
b) The length of the shorter side of the larger rectangular area is___ yd and the length of the longer side of the larger rectangular area is_____ yd.
and then
The largest total area that can be enclosed is____ yards. Thank you so much!!
Explanation / Answer
Let
length =x
width =y
since 1080 yd of fencing is available,
2x +4y =1080
x+2y= 540
x= 540 -2y
Find area as
A =xy
=(540-2y)y
=540y -2y²
To maximize find derivative and set to 0
A' = 540 -4y
540 -4y = 0
4y =540
y =540/4
y =135
hence
x = 540 -2*135 =270
Area = 270 * 135 =36450
=========
The length of the shorter side of the larger rectangular area is__135 _ yd and the length of the longer side of the larger rectangular area is__270___ yd.
and then
The largest total area that can be enclosed is_36450___ yards.
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