Premier Bank is planning to have a new building constructed. They would like the length of the building to be 40 feet longer tha
n its width. To fit their needs, each floor of the building will add 12 feet to the height of the building. The cost of the foundation of the building will be $10 per square foot. The cost for each successive floor of the building will be $35,000. They plan to spend $1,180,000 on construction. They would also like the perimeter of the building to be 3 times the height of the building. Create a system of equations to model the situation above, and use it to determine how many of the solutions are viable.
2 answers:
Let L, W, F, H, P represent, respectively, the length, width, number of floors, height, and perimeter of the building. Here are some equations that model the given information.
- L = W + 40
- H = 12×F
- foundation cost = 10×L×W
- cost of floors = 35000×F
- total cost = foundation cost + cost of floors
- total cost = 1,180,000
- P = 2×(L+W)
- P = 3×H
We can rewrite these into a system of equations involving L, W, F.
Let's look first at the relation between the perimeter and height. The two versions of perimeter are equal to each other, so we have (after substituting for H) ...
... 3H = 2(L+W)
... 3(12F) = 2((W +40) +W) . . . . substitute for H and L
... 36F = 4W +80 . . . . . . . . . . . simplify
... 9F = W +20 . . . . . . . . . . . . . divide by 4
Now, we can look at the cost equations.
... 1180000 = 35000F + 10W(W+40) . . . . . substitute for the various cost terms and for L
... 118000 = 3500F +W² +40W . . . . . . . . . divide by 10
The two simplified equations in W and F are ...
- 9F = W +20
- 3500F = 118000 -40W -W²
You can solve the first for F and substitute for F in the second. Then you have a quadratic equation in W that can be solved by the usual methods. (Likely, the quadratic formula would be the best choice.) Here, we have elected to let a graphing calculator show the two solutions. One solution has negative dimensions, so is clearly infeasible.
There is one (nearly feasible) solution:
... W ≈ 180.788
... F ≈ 22.31
There is no combination of floor dimensions and integral numbers of floors that will match <em>all</em> of the problem requirements.
_____
A solution that comes about as close as possible is ...
- length = 221 ft
- width = 181 ft
- perimeter = 804 ft
- # of floors = 22
- height = 264 ft . . . . . (3×height = 792 ft vs. 804 ft)
- cost = $1,170,010 . . . (vs. 1,180,000)
You might be interested in