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Old 03-10-2011   #2
MAS1

 
Join Date: Dec 2008
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Originally Posted by florance View Post
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.
1. The hypotenuse can be described by the line with equation:
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; 03-10-2011 at 11:50 AM.. Reason: Solution for problem 1.
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