|Jan30-12, 01:19 AM||#1|
Optimizing Dimensions to minimize Cost
1. The problem statement, all variables and given/known data
A metal can which holds 450 cm^3 is to be manufactured in cylindrical shape. The top and bottom will be cut from two squares and the corner scrap discarded (but paid for). The metal for the sides costs 0.1 cents/cm^2, while the cost for the top and bottom is 0.15cents/cm^2. Allow 0.5cents/cm for the vertical or side seam and 0.6 cents/cm for the seams joining top and bottom to the sides.
a) Determine the dimensions that would yield a minimum cost. Give dimensions to the nearest thousandth of a centimeter.
b) Determine the corresponding cost to the nearest hundreth of a cost.
2. Relevant equations
Minimizing is to find the values of a unit of function, when the tangent slope is zero.
If C'(A)=0, then we have to find the dimensions when this happens since this is when cost is minimized.
A(cylinder)=2∏r(h) + 2∏r^2
3. The attempt at a solution
I notice that part of the problem states that no matter what dimension, V=450cm^3.
since this is a can cylinder,
and so it is useful to note that:
(450cm^3)/(∏r^2)=h for all h>0 and r>0
(√((450cm^3)/(h∏)))=r for all h>0 and r>0
So, here's how I went along trying to put certain pieces together:
According to my interpretation of the problem,
Area(top + bottom)=2(∏r^2)
Cost(Area(top + bottom))=[.15cents/cm^2][2∏r^2]
Area(side)=2∏r(h) since my intuition tells me this represents the area of a cross section of the side of a cylinder where 2∏r is the circumference of the number of circles you multiply by to get the area
So also since for every radius there is a cost and every height there is a cost (I am very currently skeptical about this intuition!)
Cost(Area(side + top + bottom))= .3∏r^2[Cost(r)]^2 + .2∏r(h)[Cost(r)][Cost(h)]
Therefore, by substituting h=(450cm^3)/(∏r^2)
Cost(Area(side + top + bottom))= .3∏(r)^2[cost(r)] + .2∏r(450cm^3/∏r^2)[cost(r)][cost(h)]
Cost(Area(side + top + bottom))=.3∏r^2[(.5cent/cm)r] + .2∏r[450cm^3/∏r^2][(.5cent/cm)r]
Cost'(Area(side + top + bottom))=.9∏[.5cent/cm]r^3 + 45=0 to minimize cost, but r>0 and h>0 what am I doing wrong?
|Jan30-12, 03:42 AM||#2|
|Jan30-12, 10:03 AM||#3|
The cost of the top + bottom is (0.15)*2*(2r)^2, because you pay 0.15 per cm^2 for the whole 2r by 2r square (you pay for the discarded corners, you said).
|area, calculus 1, cylinder, fun, optimize|
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