Question

The manager of a large store wishes to add a fenced-in rectangular storage yard or 10,000 square feet, using the building as one side of the yard. Use calculus to determine the minimum amount of fencing that must be used to enclose the remaining three sides.

Answer 1

*of 10,000 feet not or.

Answer 2

I've started the problem, but I'm lost. I think that A=xy -. Area is x*y; so that would have to be equal to 10,000,

Answer 3

Area = Length X Breath
Let the Breath be x, length be y

20,000 = xy
y = 20,000/x

let fencing, P = 2x + y ( since one side of the length is the building)
P = 2x + 20,000/x
dP/dx = 2 - 20,000/x²
At turning point (P is a max or minimum), dP/dx = 0
0 = 2 - 20,000/x²
2 = 20,000/x²
2x² = 20,000
x² = 10000
x = 100 or -100(reject)
y = 20000/x
y = 200

Hence, the minimum amount of fencing is 2x + y = 2(100) + 200 = 400

Let breadth = x, length = y again
2x + y = 1600
y = 1600 - 2x

let Area, A = xy
A = x(1600 - 2x)
A = 1600x - 2x²
dA/dx = 1600 - 4x
when dA/dx = 0,
0 = 1600 - 4x
x = 400
y = 1600 - 2x = 1600 - 800 = 800

hence the largest possible yard = 400(800) = 320,000 sq ft

Answer 4

(thanks to http://answers.yahoo.com/question/index?qid=20101215030925AABMJHO)

Answer 5

That question is asking for a maximum, not a minimum.

Answer 6

Have to find the smallest amount of fencing, not the largest. ^^;