a math question?

Question

You are a manufacturer of steel containers(for liquids) of two basic shapes- tins shaped as right circular cylinders and rectangular boxes. The containers' capacity is 1 litre. Obviously you want to use the least amount of steel so as to maximise your profits. Can you work out the dimensions of the containers? How many different boxes of the same type will you have? And if possible, illustrate your response with graphs and tables, well-documented.

Answer 1

Since one litre = 1000cm^3, We can construct a cylindrical container with surface area S = 2pir^2 +2pirh.(I am using pi since I don' t have the greek symbol.) We wish to minimize S subject to the constraint that V = pir^2h = 1. Now, rewrite the equation for S as S = 2pir^2 + 2pir(1/pir^2) = 2pir^2 + 2/r). Now, differentiate S wrt r and you get ds/dr = 4pir - 2/r^2 = 0 => r = 1/pir^2. Now , use h = 1/2pir^2 to solve for h.

For the rectangular boxes, the problem is impossible without some relation between the three sides..

Let me know if you don't understand.

Answer 2

Well start out with volume and surface area formulas.

Cylinder volume is the area of the circle multiplied by the cylinder's height, pi*r^2*h. Surface area is the circulference of the circle multiplied by the height, plus twice the area of the circle (one at the top and one at the bottom), 2pi*r*h+4*pi*r^2.

Solve for r where the volume is 1L or 1000cm^3 and then find the surface area for that value of r.

Same for the cube: The volume is the area of one square multiplied by the container height, or s^3. The surface area is 6s^2.

Solve for s where the volume is 1000cm^3 and then find the surface area for that value of s.

Answer 3

You need the dimensions of the rolled steel (the raw material) before any calculations can be performed. Only then a low waste system can be designed.

Answer 4

bananas

Answer 5

i dont know
why is it so hard