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Aleksandr-060686 [28]
3 years ago
9

A solid is formed by adjoining two hemispheres to the ends of a right circular cylinder. An industrial tank of this shape must h

ave a volume of 4640 cubic feet. The hemispherical ends cost twice as much per square foot of surface area as the sides. Find the dimensions that will minimize the cost. (Round your answers to three decimal places.)
Mathematics
1 answer:
mestny [16]3 years ago
5 0

Answer:

Radius =6.518 feet

Height = 26.074 feet

Step-by-step explanation:

The Volume of the Solid formed  = Volume of the two Hemisphere + Volume of the Cylinder

Volume of a Hemisphere  =\frac{2}{3}\pi r^3

Volume of a Cylinder =\pi r^2 h

Therefore:

The Volume of the Solid formed

=2(\frac{2}{3}\pi r^3)+\pi r^2 h\\\frac{4}{3}\pi r^3+\pi r^2 h=4640\\\pi r^2(\frac{4r}{3}+ h)=4640\\\frac{4r}{3}+ h =\frac{4640}{\pi r^2} \\h=\frac{4640}{\pi r^2}-\frac{4r}{3}

Area of the Hemisphere =2\pi r^2

Curved Surface Area of the Cylinder =2\pi rh

Total Surface Area=

2\pi r^2+2\pi r^2+2\pi rh\\=4\pi r^2+2\pi rh

Cost of the Hemispherical Ends  = 2 X  Cost of the surface area of the sides.

Therefore total Cost, C

=2(4\pi r^2)+2\pi rh\\C=8\pi r^2+2\pi rh

Recall: h=\frac{4640}{\pi r^2}-\frac{4r}{3}

Therefore:

C=8\pi r^2+2\pi r(\frac{4640}{\pi r^2}-\frac{4r}{3})\\C=8\pi r^2+\frac{9280}{r}-\frac{8\pi r^2}{3}\\C=\frac{9280}{r}+\frac{24\pi r^2-8\pi r^2}{3}\\C=\frac{9280}{r}+\frac{16\pi r^2}{3}\\C=\frac{27840+16\pi r^3}{3r}

The minimum cost occurs at the point where the derivative equals zero.

C^{'}=\frac{-27840+32\pi r^3}{3r^2}

When \:C^{'}=0

-27840+32\pi r^3=0\\27840=32\pi r^3\\r^3=27840 \div 32\pi=276.9296\\r=\sqrt[3]{276.9296} =6.518

Recall:

h=\frac{4640}{\pi r^2}-\frac{4r}{3}\\h=\frac{4640}{\pi*6.518^2}-\frac{4*6.518}{3}\\h=26.074 feet

Therefore, the dimensions that will minimize the cost are:

Radius =6.518 feet

Height = 26.074 feet

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A food-packaging apparatus underfills 10% of the containers. Find the probability that for any particular 10 containers the numb
Maksim231197 [3]

Answer:

a) P(X = 1) = 0.38742

b) P(X = 3) = 0.05740

c) P(X = 9) = 0.00000

d) P(X \geq 5) = 0.00163

Step-by-step explanation:

For each container, there are only two possible outcomes. Either it is undefilled, or it is not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem

There are 10 containers, so n = 10.

A food-packaging apparatus underfills 10% of the containers, so p = 0.1.

a) This is P(X = 1)

P(X = 1) = C_{10,1}.(0.1)^{1}.(0.9)^{9} = 0.38742

b) This is P(X = 3)

P(X = 3) = C_{10,3}.(0.1)^{3}.(0.9)^{7} = 0.05740

c) This is P(X = 9)

P(X = 9) = C_{10,9}.(0.1)^{9}.(0.9)^{1} = 0.00000

d) This is P(X \geq 5).

Either the number is lesser than five, or it is five or larger. The sum of the probabilities of each event is decimal 1. So:

P(X < 5) + P(X \geq 5) = 1

P(X \geq 5) = 1 - P(X < 5)

In which

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{10,0}.(0.1)^{0}.(0.9)^{10} = 0.34868

P(X = 1) = C_{10,1}.(0.1)^{1}.(0.9)^{9} = 0.38742

P(X = 2) = C_{10,2}.(0.1)^{2}.(0.9)^{8} = 0.1937

P(X = 3) = C_{10,3}.(0.1)^{3}.(0.9)^{7} = 0.05740

P(X = 4) = C_{10,4}.(0.1)^{1}.(0.9)^{9} = 0.38742

So

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.34868 + 0.38742 + 0.19371 + 0.05740 + 0.01116 = 0.99837

Finally

P(X \geq 5) = 1 - P(X < 5) = 1 - 0.99837 = 0.00163

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