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Tasya [4]
2 years ago
14

A microwaveable cup-of-soup package needs to be constructed in the shape of cylinder to hold 550 cubic centimeters of soup. The

sides and bottom of the container will be made of styrofoam costing 0.04 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 cost.
Helpful information: h : height of cylinder, r : radius of cylinder
Volume of a cylinder: V = pir²h Area of the sides: A = 2pirh Area of the top/bottom: A = pir²
Physics
1 answer:
myrzilka [38]2 years ago
7 0

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.

Learn more about radius at

brainly.com/question/24375372

#SPJ4

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Complete question:

At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10​⁵​​ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.

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Explanation:

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Also;

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t = \frac{M}{qB} = \frac{9.11 *10^{-31}}{1.602*10^{-19}*3.47*10^{-3}} = 1.639 *10^{-9}  \ s \ = 1.639 \ ns

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.

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