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Vaselesa [24]
2 years ago
12

A manufacturer wants to make open tin boxes from pieces of tin with dimensions 8 in. by 15 in. by cutting equal squares from the

four corners and turning up the sides. Find the side of the square cutout that gives the box the largest possible volume.
0 in.
6 in.
52788.gif
52789.gif
Mathematics
1 answer:
Stells [14]2 years ago
5 0

Answer:

\frac{5}{3} in

Step-by-step explanation:

Let x be the side of square.

Length of box=8-2x

Width of box=15-2x

Height of box=x

Volume of box=L\times B\times H

Substitute the values then we get

Volume of box=V(x)=(8-2x)(15-2x)x=(15x-2x^2)(8-2x)

V(x)=120x-30x^2-16x^2+4x^3

V(x)=4x^3-46x^2+120x

Differentiate w.r.t x

V'(x)=12x^2-92x+120

V'(x)=0

12x^2-92x+120=0

3x^2-23x+30=0

3x^2-18x-5x+30=0

3x(x-6)-5(x-6)=0

(x-6)(3x-5)=0

x-6=0\implies x=6

3x-5=0\implies x=\frac{5}{3}

Again differentiate w.r.t x

V''(x)=24x-92

Substitute x=6

V''(6)=24(6)-92=52>0

Substitute x=5/3

V''(5/3)=24(5/3)-92=-52

Hence, the volume is maximum at x=\frac{5}{3}

Therefore, the side of the square ,x=\frac{5}{3} in cutout that gives the box the largest possible volume.

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1E18 OR 10^18

Step-by-step explanation:

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If oranges cost $0.75 for 1/3 of a dozen, how many oranges can you buy for $3.00?? ❗️Please help❗️
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√17, 4.23, 9/2

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--------------------
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5 0
3 years ago
New York City is the most expensive city in the United States for lodging. The mean hotel room rate is per night (USA Today, Apr
kow [346]

Answer:

P(X

Step-by-step explanation:

Assuming a mean of $204 per night and a deviation of $55.

a. What is the probability that a hotel room costs $225 or more per night (to 4 decimals)?

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean"

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Let X the random variable that represent the cost per night at the hotel, and for this case we know the distribution for X is given by:

X \sim N(204,55)  

Where \mu=204 and \sigma=55

And let \bar X represent the sample mean, the distribution for the sample mean is given by:

\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})

We are interested on this probability

P(X>225)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X>225)=P(\frac{X-\mu}{\sigma}>\frac{225-\mu}{\sigma})

=P(Z>\frac{225-204}{55})=P(Z>0.382)

And we can find this probability on this way:

P(Z>0.382)=1-P(Z

8 0
2 years ago
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