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sveta [45]
3 years ago
6

Six identical squares are cut from the corners and edges of an 80 cm by 50 cm cardboard rectangle. the remaining piece is folded

into a closed box with 3 flaps. what is the largest possible volume of such a box?

Mathematics
2 answers:
swat323 years ago
7 0

Solution:

The Given Rectangle having dimensions

Length = 80 cm

Breadth = 50 cm

Let the Six squares which has been cut from this rectangle have side of length a cm.

Area of each square = (Side)²= a²

Area of 6 Identical Squares = 6 × a²= 6 a²

If four squares are cut from four corners and two along Length,

then , Length of Box = (80 - 3 a)cm, Breadth of Box = (50 - 2 a)cm, Height = a cm

Volume of Box =V = Length × Breadth × Height

    V = (80 - 3 a)× (50 - 2 a)× a

     V   = 4000 a - 310 a² + 6 a³

For Maximum Volume

V'= 0 , where V' = Derivative of V with respect to a.

V'= 4000 - 620 a + 18 a²

V' =0

18 a² - 620 a + 4000= 0

9 a² - 310 a + 2000=0

using determinant method

a = \frac{310\pm\sqrt{96100-72000}}{18}=\frac{310\pm155}{18}=\frac{310-155}{18}=8.6 cm

V"=1 8 a - 310 = -ve

which shows , when a = 8.6 cm , volume is maximum.

So, V = 4000×8.6 - 310×(8.6)²+6×(8.6)³=15288.736 cm³

OR

If four squares are cut from four corners and two along Breadth,

then , Length of Box = (80 - 2 a)cm, Breadth of Box = (50 -  3 a)cm, Height = a cm

Volume of Box =V = Length × Breadth × Height

    V = (80 - 2 a)× (50 - 3 a)× a

     V   = 4000 a - 340 a² + 6 a³

For Maximum Volume

V'= 0 , where V' = Derivative of V with respect to a.

V'= 4000 - 680 a + 18 a²

V' =0

18 a² - 680 a + 4000= 0

9 a² - 340 a + 2000=0

Using Determinant method

a = \frac{340\pm\sqrt{115600-72000}}{18}=\frac{340\pm209}{18}=\frac{131}{18}=7.6 cm

V"=18 a -340= -ve value when a = 7.6 cm, shows volume is maximum when a = 7.6 cm

V= 4000×7.6 -340 × (7.6)² +6× (7.6)³=13395.456 cubic cm

Anna11 [10]3 years ago
5 0
Check the picture.

let the length of a side of each of the squares removed be x.

The box formed will have dimensions: 80-2x, 50-2x, x(the height)

So the volume can be expressed as a function of x as follows:

f(x)=(80-2x)(50-2x)x=[4000-160x-100x+4 x^{2} ]x=(4 x^{2}-260x+4000)x

so f(x)=4 x^{3}-260x^{2}+4000x

the solutions of f'(x)=0 gives the inflection points, so the candidates for maxima points,

f'(x)=12x^{2}-520 x +4000=0

solving the quadratic equation, either by a calculator, graphing software, or by other algebraic methods as the discriminant formula, we find the solutions

x=10 and x=33.333

plug in f(x) these values to see which greater:

f(10)=(80-20)(50-20)10=60*30*10=18000 cm cubed

f(33.333)=(80-66.666)(50-66.666)33.333= which is negative because (50-66.666)<0



Answer: 18000 cm cubed

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3 years ago
A cylinder has a base radius of 4m and a height of 6m. What is its volume in cubic m, to the nearest tenths place?
Ronch [10]
<h3>Answer:</h3>

301.6 cubic meters

<h3>Step-by-step explanation:</h3>

A cylinder is a shape with straight sides with circular or oval cross-sections. We know that the cylinder in the question must be a circular cylinder due to its radius description. 

Volume Formula

A circular cylinder has a volume of V=\pi r^2h. In this equation, V is the volume, r is the radius, and h is the height. The question tells us that r=4m and h=6m. So, we can plug these values into the formula.

Solving for Volume

To solve plug the values into the formula and rewrite the equation.

  • V=\pi *4^2*6

Next, apply the exponents.

  • V=\pi *16*6

Then, multiply the constants.

  • V=\pi *96

Finally, multiply the remaining terms. Remember to use the pi button on the calculator and not an estimation to get a more exact value.

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Make sure your answer is rounded to the correct digit. This means that the volume must be 301.6 cubic meters.

6 0
2 years ago
Written form of 73,046
Helen [10]
73,046 in written form is seventy three thousand forty six
6 0
3 years ago
Read 2 more answers
1. Find the length of X (in the picture) plssss I need help.​
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Step-by-step explanation:

\frac{5}{4}  =  \frac{x}{6}  \\ 4x = 30 \\ x = 7.5

3 0
3 years ago
Let X1 and X2 be independent random variables with mean μand variance σ².
My name is Ann [436]

Answer:

a) E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu

So then we conclude that \hat \theta_1 is an unbiased estimator of \mu

E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu

So then we conclude that \hat \theta_2 is an unbiased estimator of \mu

b) Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}

Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}

Step-by-step explanation:

For this case we know that we have two random variables:

X_1 , X_2 both with mean \mu = \mu and variance \sigma^2

And we define the following estimators:

\hat \theta_1 = \frac{X_1 + X_2}{2}

\hat \theta_2 = \frac{X_1 + 3X_2}{4}

Part a

In order to see if both estimators are unbiased we need to proof if the expected value of the estimators are equal to the real value of the parameter:

E(\hat \theta_i) = \mu , i = 1,2

So let's find the expected values for each estimator:

E(\hat \theta_1) = E(\frac{X_1 +X_2}{2})

Using properties of expected value we have this:

E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu

So then we conclude that \hat \theta_1 is an unbiased estimator of \mu

For the second estimator we have:

E(\hat \theta_2) = E(\frac{X_1 + 3X_2}{4})

Using properties of expected value we have this:

E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu

So then we conclude that \hat \theta_2 is an unbiased estimator of \mu

Part b

For the variance we need to remember this property: If a is a constant and X a random variable then:

Var(aX) = a^2 Var(X)

For the first estimator we have:

Var(\hat \theta_1) = Var(\frac{X_1 +X_2}{2})

Var(\hat \theta_1) =\frac{1}{4} Var(X_1 +X_2)=\frac{1}{4} [Var(X_1) + Var(X_2) + 2 Cov (X_1 , X_2)]

Since both random variables are independent we know that Cov(X_1, X_2 ) = 0 so then we have:

Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}

For the second estimator we have:

Var(\hat \theta_2) = Var(\frac{X_1 +3X_2}{4})

Var(\hat \theta_2) =\frac{1}{16} Var(X_1 +3X_2)=\frac{1}{4} [Var(X_1) + Var(3X_2) + 2 Cov (X_1 , 3X_2)]

Since both random variables are independent we know that Cov(X_1, X_2 ) = 0 so then we have:

Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}

7 0
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