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03092011  #1 
Join Date: Mar 2011
Posts: 2

Calculus optimization problems 2
1. determine tha area of the largest rectangle that can be inscribed in a right triangle if the legs adjenct to the right angles are 5 cm and 12 cm long. teh two sides of the rectangle lie along the legs.
2 A cylindrical shaped tin can must have a Vol = 1000 cm3 .determine the dimensions that require the minimum ammount of tin for the can. According to the marketing department the smallest can that the market will accept has a diameter of 6 cm and a height of 4 cm. b. express your answer for part 1 as a ratio of height to diameter. Does this ratio meet the requirements outline by marketing depart. 
03102011  #2  
Join Date: Dec 2008
Posts: 249

Quote:
y = (12/5)x + 12 Area = xy = x((12/5)x + 12) Area = (12/5)x^2 + 12x Take derivative of area, set equal to 0, and solve for x. der(Area) = (24/5)x + 12 0 = (24/5)x + 12 x = 5/2 = 2.5 cm y = (12/5)(5/2) + 12 = 6 cm Area = 15 cm^2 2.A. volume of cylinder = pi*(r^2)*h = 1000 cm^3 Then h = 1000/(pi*r^2) Amount of tin for the can is the surface area of a cylinder given by: S.A. = area of top + area of bottom + area of side S.A. = pi*r^2 + pi*r^2 + 2*pi*r*h S.A. = 2*pi*r^2 + 2*pi*r*h Substituting for h gives: S.A. = 2*pi*r^2 + 2*pi*r*1000/(pi*r^2) S.A. = 2*pi*r^2 + 2000/r To minimize the surface area, take the derivative (with respect to r) and set it equal to 0. derivative(S.A.) = 4*pi*r + (2000/r^2) 0 = 4*pi*r  2000/r^2 0 = (4*pi*r^3  2000)/r^2 4*pi*r^3 = 2000 r^3 = 500/pi r^3 = 159.155... r = 5.419 cm (diam = 10.838 cm) h = 1000/(pi*5.419^2) = 10.838 cm 2.B. h/d = 10.838/10.838 = 1 Last edited by MAS1; 03102011 at 11:50 AM.. Reason: Solution for problem 1. 

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