I am sooo lost please help!!!!?

Question

Plans for a new supermarket require a floor area of 14,400 ft². The supermarket is to be rectangular in shape with three solid brick walls and a very fancy all-glass front. If glass cost 1.88 times as much as the brick wall per linear foot, what should be the dimensions of the building so that the cost of material for the walls is a minimum?

There are four walls 3 brick and 1 glass.

Answer 1

Thinking about this intuitively, if the cost for all 4 walls was the same, what design would give you the smallest perimeter of a rectangle where the product of the sides is 14,400? That would be a square. Now, you can see that because that one wall has a greater cost, you want to make that wall's length less (which means making the corresponding back wall less, and the sides grow longer). But at some point we lose so much efficiency (because the side walls get SO much longer that the total price goes up, that we start paying more. We need to figure out at what point the cost of the front + back walls equals the cost of the side walls.

Let's develop a couple of formulas.

let c = the cost/linear foot, x = length of one wall (we will arbitrarily assume that x and it's partner x' will contain the glass), y = length of an adjacent wall.

Then total cost = c*x + 1.88*c*x +2*c*y or simply 2.88*c*x + 2*c*y

The total square footage of 14400 is equal to x * y: x * y = 14400

Since we already observed that the greatest economy is found when the cost of the front and back walls (2.88*c*x) is equal to the cost of the side walls (2*c*y) we can solve this equation:

2.88*c*x = 2*c*y; or 1.44*x = y

substitute for y in: x * y = 14400 yields:

x * 1.44*x = 14400, which simplifies as:

1.44*(x^2) = 14400, which when solved yields:

x = 100 feet

This implies that y = 144 feet.

Answer 2

Let x be the length of the glass front in feet. Then x is also be the length of one brick wall. Let y be the common length of the other two brick walls.
The total cost is 2.88x + 2y times the cost of a foot of brick wall. The supermarket's area is x*y = 14,400, that is, y = 14,400/x. So the problem is to minimize

2.88x + 2*14400/x.

Differentiate w.r.t. x, and set equal to 0:

0 = (d/dx)(2.88x + 2*14400/x) = 2.88 - 28800/x^2

x^2 = 10000

x = 100
y = 14400/x = 144

That is, the glass front should be 100 ft long, and the other side length should be 144 ft.

Answer 3

L x W = 14400

L = 14400/W

3 walls (2L and 1W) will be brick. 1 wall will be glass. Let's call the cost of a brick wall 1 and the cost of a glass wall 1.88. For purpose of this analysis, that will work since it reflects the proper proportion Call the height H, which is some fixed constant.

Cost = C = 2LH(1) + WH (1) + WH(1.88)

substitute 14400/W for L

C = 28800H/W + 2.88WH

C' = -28800H/(W^2) + 2.88H = 0

28800/(W^2) = 2.88

W^2 = 28800/2.88 = 10000

W = 100, L = 144

So 100 ft x 144 ft will minimize cost.

Answer 4

Easy. One foot wide (or narrower) x 14,440 feet longer (or longer), with one one foot wall having the glass in it.

Obviously your question does not have enough info in it to make sense, or this non-workable answer would not be true. There are always 4 walls in a rectangular building, and your problem assumes 3. A wall of all windows is still a wall. Therefore the question is fatally flawed.

Show this proof to your instructor tomorrow, and you should receive an A. Then again,, s/he may have a closed mind & call you a smart ______. !!

Answer 5

Depends on the minimum & maximum width & length permissible.

The more narrow the width of the building in this case,
the less the cost of the walls; but then if the building is, say, only 1 ft. wide, it will be the cheapest to build but utterly useless, to take an extreme example.

;-)