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almond37 [142]
4 years ago
11

An open box is constructed of 3500 cm2 of cardboard. The box is a cuboid, with height hcm and square base of side xcm. What is t

he value of x which maximises the volume of the box?
Mathematics
1 answer:
erica [24]4 years ago
7 0

Answer:

34.16 cm

Step-by-step explanation:

side of square base = x

height = h

area, A = 3500 cm^2

Area = x² + 4xh = 3500

4 xh = 3500 - x²

h = (3500 - x²)/4x

Volume = Area of base x height

V = x² h

V = x² (3500 - x²)/4x

V = (3500 x - x³) / 4

Differentiate volume with respect to x

dV/dx = (3500 - 3x²) / 4

It is equal to zero for maxima and minima

3500 - 3x² = 0

x = 34.16 cm

Now differentiate again

d²V/dx² = 6x / 4

It is negative so the volume is maximum.

Thus, for x = 34.16 cm, the volume is maximum.

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taurus [48]

Answer:

a) We need 4 stations to be 98% certain that an enemy plane flying over will be detected by at least one station.

b) If seven stations are in use, the expected number of stations that will detect an enemy plane is 4.55.

Step-by-step explanation:

For each station, there are two two possible outcomes. Either they detected the enemy plane, or they do not. This means that we can solve this problem using concepts of 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, we have that:

The probability that a single radar station will detect an enemy plane is 0.65. This means that n = 0.65.

(a) How many such stations are required to be 98% certain that an enemy plane flying over will be detected by at least one station?

This is the value of n for which P(X = 0) \leq 0.02.

n = 1.

P(X = 0) = C_{1,0}.(0.65)^{0}.(0.35)^{1} = 0.35

n = 2

P(X = 0) = C_{2,0}.(0.65)^{0}.(0.35)^{2} = 0.1225

n = 3

P(X = 0) = C_{3,0}.(0.65)^{0}.(0.35)^{3} = 0.0429

n = 4

P(X = 0) = C_{4,0}.(0.65)^{0}.(0.35)^{4} = 0.015

We need 4 stations to be 98% certain that an enemy plane flying over will be detected by at least one station.

(b) If seven stations are in use, what is the expected number of stations that will detect an enemy plane?

The expected number of sucesses of a binomial variable is given by:

E(x) = np

So when n = 7

E(x) = 7*(0.65) = 4.55

If seven stations are in use, the expected number of stations that will detect an enemy plane is 4.55.

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Step-by-step explanation:

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