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umka21 [38]
3 years ago
7

. Aladdin found a cave filled with boxes of gold, measuring 2 inches × 2 inches × 2 inches, and boxes of silver, measuring 1 inc

h × 1-inch × 1 inch. However, Aladdin brought only one box measuring 6 inches × 6 inches × 7 inches with him. He filled it with as many boxes of gold as would fit, then put as many boxes of silver as would fit in the remaining space. How many boxes of treasure did Aladdin take?
Mathematics
1 answer:
Viktor [21]3 years ago
4 0

Answer:

Step-by-step explanation:

volume of box=6×6×7 in³

he fits gold 2×2×2

so he fits it in 6×6×6

so number of gold boxes=(6×6×6)/(2×2×2)=27

remaining space=1×6×6

number of silver boxes=(1×6×6)/(1×1×1)=36

total boxes he took=27+36=63

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a_sh-v [17]
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2 years ago
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Please explain this problem!!!​
9966 [12]

tis a little of plain differentiation.

we know the radius of the cone is decreasing at 10 mtr/mins, or namely dr/dt = -10, decreasing, meaning is negative.

we know the volume is decreasing at a rate of 1346 mtr/mins or namely dV/dt = -1346, also negative.

so, when h = 9 and V = 307, what is dh/dt in essence.

we'll be needing the "r" value at that instant, so let's get it

V=\cfrac{1}{3}\pi r^2 h\implies 307=\cfrac{\pi }{3}r^2(9)\implies \sqrt{\cfrac{307}{3\pi }}=r

now let's get the derivative of the volume of the cone

V=\cfrac{1}{3}\pi r^2 h\implies \cfrac{dV}{dt}=\cfrac{\pi }{3}\stackrel{product~rule}{ \left[ \underset{chain~rule}{2r\cdot \cfrac{dr}{dt}}\cdot h+r^2\cdot \cfrac{dh}{dt} \right]} \\\\\\ -1346=\cfrac{\pi }{3}\left[2\sqrt{\cfrac{307}{3\pi }}(-10)(9)~~+ ~~ \cfrac{307}{3\pi } \cdot \cfrac{dh}{dt}\right]

-\cfrac{4038}{\pi }=-\cfrac{180\sqrt{307}}{\sqrt{3\pi }}+\cfrac{307}{3\pi } \cdot \cfrac{dh}{dt}\implies \left[ -\cfrac{4038}{\pi }+\cfrac{180\sqrt{307}}{\sqrt{3\pi }} \right]\cfrac{3\pi }{307}=\cfrac{dh}{dt} \\\\\\ -\cfrac{12114}{307}+\cfrac{180\sqrt{3\pi }}{\sqrt{307}}=\cfrac{dh}{dt}\implies -7.920939735970634 \approx \cfrac{dh}{dt}

5 0
2 years ago
A cardboard box without a lid is to have a volume of 8,788 cm3. Find the dimensions that minimize the amount of cardboard used.
Montano1993 [528]

Answer:

  x = y = 26 cm; z = 13 cm

Step-by-step explanation:

The generic solution for the minimum material in an open-top box is that the box is square and half as high as it is wide. It is half a cube of twice the volume.

The dimensions of the square base are ...

  ∛(2·8788 cm³) = 26 cm

Then the height is half that, or 13 cm.

  x = y = 26 cm; z = 13 cm

_____

If you need to see the development, you can use the method of Lagrange multipliers to find the minimum area for the given volume;

  area = xy +2(xz +yz)

  volume = xyz = 8788

We require each of the partial derivatives of L with respect to x, y, z, and λ to be zero.

  L = xy +2(xz +yz) +λ(xyz -8788)

  partial with respect to x: 0 = y+2z +λyz

  partial with respect to y: 0 = x +2z +λxz

  partial with respect to z: 0 = 2x+2y +λxy

  partial with respect to λ: 0 = xyz -8788

From the first two equations, we have ...

  λ = (y +2z)/(yz) = 1/z +2/y

  λ = (x +2z)/(xz) = 1/z +2/x

Equating these expressions for λ, we find ...

  1/z +2/y = 1/z +2/x   ⇒   x = y

The third equation then tells us ...

  λ = (2x +2y)/(xy) = 2/y +2/x

Comparing this to either of the first two expressions for λ, we see ...

  1/z +2y = 2/x +2/y   ⇒   z = x/2

This is the result we started the answer with:

  x = y = 2z = ∛(2·8788 cm³)

7 0
2 years ago
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aksik [14]

Answer:

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Step-by-step explanation:

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2 years ago
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3 years ago
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