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3 years ago

What are the dimensions of a box that would hold 250 cubic centimeters of juice and have a minimum surface area

1 answer:

3 0

The dimensions of a box that have the minium surface area for a given Volume is such that it is a cube. This is the three dimensions are equal:

V = x*y*z , x=y=z => V = x^3, that will let you solve for x,

x = ∛(V) = ∛(250cm^3) = 6.30 cm.

Answer: 6.30 cm * 6.30cm * 6.30cm. This is a cube of side 6.30cm.

The demonstration of that the shape the minimize the volume of a box is cubic (all the dimensions equal) corresponds to a higher level (multivariable calculus).

I guess it is not the intention of the problem that you prove or even know how to prove it (unless you are taking an advanced course).

Nevertheless, the way to do it is starting by stating the equations for surface and apply two variable derivation to optimize (minimize) the surface.

You do not need to follow with next part if you do not need to understand how to show that the cube is the shape that minimize the surface.

If you call x, y, z the three dimensions, the surface is:

S = 2xy + 2xz + 2yz (two faces xy, two faces xz and two faces yz).

Now use the Volumen formula to eliminate one variable, let's say z:

V = x*y*z => z = V /(x*y)

=> S = 2xy + 2x [V/(xy)[ + 2y[V/(xy)] = 2xy + 2V/y + 2V/x

Now find dS, which needs the use of partial derivatives. It drives to:

dS = [2y - 2V/(x^2)] dx + [2x - 2V/(y^2) ] dy = 0

By the properties of the total diferentiation you have that:

2y - 2V/(x^2) = 0 and 2x - 2V/(y^2) = 0

2y - 2V/(x^2) = 0 => V = y*x^2

2x - 2V/(y^2) = 0 => V = x*y^2

=> y*x^2 = x*y^2 => y*x^2 - x*y^2 = xy (x - y) = 0 => x = y

Now that you have shown that x = y.

You can rewrite the equation for S and derive it again:

S = 2xy + 2V/y + 2V/x, x = y => S = 2x^2 + 2V/x + 2V/x = 2x^2 + 4V/x

Now find S'

S' = 4x - 4V/(x^2) = 0 => V/(x^2) = x => V =x^3.

Which is the proof that the box is cubic.

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3 years ago

Together, Dante and Mia have a total of 350 pennies in their piggy banks. After Dante lost ½ of his pennies and Mia lost ⅓ of he

natta225 [31]

Answer:

Altogether they lost 150 pennies.

Step-by-step explanation:

X = Dante's pennies

Y = Mia's pennies

X + Y = 350 1/2 X = 2/3 Y

X = 4/3 Y

4/3 Y + Y = 350

7/3 Y = 350

Y = 350 * 3/7

Y = 50 * 3

Y = 150

X + 150 = 350

X = 200

1/2 * 200 = 2/3 * 150

100 . = 100

1/2 * 200 = 100 -> Dante lost 100 pennies

1/3 * 150 = 50 -> Mia lost 50 pennies

Altogether they lost 150 pennies.

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What is the answer to 7/10+ 1/2

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Answer:

12/10 or 1 1/5

Step-by-step explanation:

1/2 = 5/10

7/10 + 5/10 = 12/10 or 1 1/5

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2 years ago

a) What is an alternating series? An alternating series is a whose terms are__________ . (b) Under what conditions does an alter

andriy [413]

Answer:

a) An alternating series is a whose terms are alternately positive and negative

b) An alternating series $\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} (-1)^{n-1} b_n$ where bn = |an|, converges if $0< b_{n+1} \leq b_n$ for all n, and $\lim_{n \to \infty} b_n =$ 0

c) The error involved in using the partial sum sn as an approximation to the total sum s is the remainder Rn = s − sn and the size of the error is bn + 1

Step-by-step explanation:

Part a

An Alternating series is an infinite series given on these three possible general forms given by:

$\sum_{n=0}^{\infty} (-1)^{n} b_n$

$\sum_{n=0}^{\infty} (-1)^{n+1} b_n$

$\sum_{n=0}^{\infty} (-1)^{n-1} b_n$

For all $a_n >0, \forall n$

The initial counter can be n=0 or n =1. Based on the pattern of the series the signs of the general terms alternately positive and negative.

Part b

An alternating series $\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} (-1)^{n-1} b_n$ where bn = |an| converges if $0< b_{n+1} \leq b_n$ for all n and $\lim_{n \to \infty} b_n$ =0

Is necessary that limit when n tends to infinity for the nth term of bn converges to 0, because this is one of two conditions in order to an alternate series converges, the two conditions are given by the following theorem:

Theorem (Alternating series test)

If a sequence of positive terms {bn} is monotonically decreasing and

$\lim_{n \to \infty} b_n = 0$ , then the alternating series $\sum (-1)^{n-1} b_n$ converges if:

i) $0 \leq b_{n+1} \leq b_n \forall n$

ii) $\lim_{n \to \infty} b_n = 0$

then $\sum_{n=1}^{\infty}(-1)^{n-1} b_n$ converges

Proof

For this proof we just need to consider the sum for a subsequence of even partial sums. We will see that the subsequence is monotonically increasing. And by the monotonic sequence theorem the limit for this subsquence when we approach to infinity is a defined term, let's say, s. So then the we have a bound and then

$|s_n -s| < \epsilon$ for all n, and that implies that the series converges to a value, s.

And this complete the proof.

Part c

An important term is the partial sum of a series and that is defined as the sum of the first n terms in the series

By definition the Remainder of a Series is The difference between the nth partial sum and the sum of a series, on this form:

Rn = s - sn

Where $s_n$ represent the partial sum for the series and s the total for the sum.

Is important to notice that the size of the error is at most $b_{n+1}$ by the following theorem:

Theorem (Alternating series sum estimation)

If $\sum (-1)^{n-1} b_n$ is the sum of an alternating series that satisfies

i) $0 \leq b_{n+1} \leq b_n \forall n$

ii) $\lim_{n \to \infty} b_n = 0$

Then then $\mid s - s_n \mid \leq b_{n+1}$

Proof

In the proof of the alternating series test, and we analyze the subsequence, s we will notice that are monotonically decreasing. So then based on this the sequence of partial sums sn oscillates around s so that the sum s always lies between any two consecutive partial sums sn and sn+1.

$\mid{s -s_n} \mid \leq \mid{s_{n+1} -s_n}\mid = b_{n+1}$

And this complete the proof.

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3 years ago