FPM 20130301  | Scoins.net | DJS

## FPM 20130301 

More of the same, but most of a month later. Class indolent but, they say, working on other subjects. I do hope so. Meanwhile the generation of questions continues and I’m capturing 25% of them or less to put here.

DJS 20130312

FPM.
Calculus, differentiation, uses of:

1.  A large square sheet of metal, area 144dm² is used to create an open box with a square base of side a. To do this, squares of side x are cut from the sheet, which is then folded and welded. Write expressions for the area, A, and the volume, V, of the resulting shape. Hence discover the dimensions that give maximum and minimum volume.

2.  A sealed cylindrical can, height h=kr and radius r, is to be made from a fixed area, A. Find the value of k that gives the maximum volume.

3.  An open-topped cylinder has an area of 1 square metre. Write an expression for the area in terms of height and radius, h and r respectively. Find the ratio h/r that gives a maximum volume.

4. A hopper is to be made from sheet metal in the form of a right circular cone, apex downwards, height h and radius r. The cone is open at the top. Write expressions for surface area and volume and hence show that the ratio of height to radius that will give maximum volume for a fixed area is    h = r √2.

5. Rectangular panels are used to create an open cuboid tank for holding oil. Each panel has dimension a and b, in metres; assume b>a. The resulting tank has volume V. The cost of a linear joint is p per metre, and the cost of cutting a sheet is 2p per metre. There are two tank styles for sale; one, Model S, with a square base and one, Model R, with a rectangular base, each of volume a²b. The cost of a tank is the construction cost plus the material cost, where unused material can be returned to store.
(i) Draw diagrams to show both styles.
(ii) Show that style S has a construction cost CCS = (6a + 4b)p (where p is the cost per metre of a joint) and write a corresponding expressions for the cost of constructing tanks of style R.
iii) What ratio would a and b have for the construction costs to be the same for R and S?
iv) If the sheet cost is k p/sq metre then show that, for the model S tank, tank cost, TC, is TC(S) = (6a + 4b + k(a
²+4ab))p and write the corresponding expression for TC(R). Rewrite these for the special case where ak=5.
v) For a fixed size, V=10, and the special case where ak = 5, find a stationary value. Show that for Model S this is a minimum close to a=2m; find the equivalent position for model R, explain what has happened to the model and hence find an appropriate conclusion for model R
vi) We stated that b>a; for V=10, what limitations does this put on a and b?
vii) Clearly a,b,k are all >0; show that the equality TCS=TCR reduces to  a(4+ak) = b (2+ak). Show that this reveals both that 2a>b and that (1+ak)/(2+ak)>0.25. Does either inequality add information?
viii) What happens when a=b?

Q1 Minimum at x=6, V=0 maximum at x=2, V= 2.8.8 = 128 dm3

Q2: Until someone disagrees, I think:  (iv) TCS = (11a+24b)p, TCR = (20a +17b)p.

(v) for S: a³ = 480/11 => a=3.52; b = 0.81; TC(S)min = 58p. For R: a³= 17 => a=2.57; b = 1.51; TC(R)min = 77p. Note: a>b here, so reject the solution. One response is that model R should be treated as a cube [consequence; a=b=10 1/3, TCR/p =8a+5a(ak)=33k=71.1]; another is that ak≠5 for model R, more like ak=1, TCR=42.8.(vi) a3 <10, 0<a<2.154, b>2.154

Q3: several ways to do this, dV/dr and dV/dk; maximum volume at k=2.

Q4: h=r; Qr h=r √2

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