A microwaveable cup-of-soup package needs to be constructed in the shape of a cylinder to hold 600 cubic centimeters of soup. The sides and bottom of the container will be made of styrofoam costing 0.02 cents per square centimeter. The top will be made of glued paper, costing 0.05 cents per square centimeter. Find the dimensions for the package that will minimize production costs.
h: height of the cylinder, r: radius of the cylinder
The volume of a cylinder: V=πr2h
Area of the sides: A=2πrh
Area of the top/bottom: A=πr2
The cost of packaging, C=2πrh*0.02+ πr^2*0.02+ πr^2*0.05 subject to the constraint πr^2h=600
C=πr(0.04h+.07r) and the constraint implies h=600/ πr^2
So C=πr(24/πr^2+.07r)=24/r+.07πr^2
C'=-24/r^2+0.14πr=0
r^3=24/0.14π r=3.79 cm
h=600/πr^2=13.3 cm
C=π*3.79*(0.04*13.3+.07*3.79)=9.48cents
C''=0.14π+48/r^3>0 for all r>=0 so our solution is indeed a minimum.
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If the crate is moving along the floor in the same direction with a constant speed, it is in dynamic equilibrium. Equilibrium means there is no net force acting on the crate.
Since there is no net force on the crate, there must be a friction force on the crate equal in magnitude and opposite in direction to the applied horizontal force. Therefore the force of friction acting on the crate is 100N.
B.) acceleration.
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Answer: 4.9 x 10-3 N
Explanation:
A = 500cm^2 = 5 x 10^-2 m^2
V = 5 m/s
R = 10^-3 g/cm^2.sec = 10^-2kg/m^2 . sec
Prain water = R / V = 10^-2 / 5 = 2 x 10-3 kg/m^3
For the stationary bowl,
dm/dt =pAv= RA
F= dp/dt = (dm/dt) v = RAv = 2.5 x 10^-3 N
Bowl moving upwards to speed u = 2 m/s
dm/dt = pA ( v + u) / v
F = dp/dt = (dm/dt)(v+u) = RA (v+u)^2 / v = 4.9 x 10^-3 N
One atom on nitrogen because there isn’t a number directly after it’s the 3 signifies that there is three groups of those atoms as a total
