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stealth61 [152]
3 years ago
14

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.

(Let x, y, and z be the dimensions of the cardboard box.)
Mathematics
1 answer:
Montano1993 [528]3 years ago
7 0

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)

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The graph of which function will contain the point(2,10)​
kotegsom [21]

Answer:

point (2 , 10)

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3 0
3 years ago
Y=x^2-5x+11<br> Find vertex form
Marat540 [252]

Answer:

42

Step-by-step explanation:

The answer to life the universe and everything is 42

8 0
3 years ago
How do you solve this?
solong [7]

It looks like your equations are

7M - 2t = -30

5t - 12M = 115

<u>Solving by substitution</u>

Solve either equation for one variable. For example,

7M - 2t = -30   ⇒   t = (7M + 30)/2

Substitute this into the other equation and solve for M.

5 × (7M + 30)/2 - 12M = 115

5 (7M + 30) - 24M = 230

35M + 150 - 24M = 230

11M = 80

M = 80/11

Now solve for t.

t = (7 × (80/11) + 30)/2

t = (560/11 + 30)/2

t = (890/11)/2

t = 445/11

<u>Solving by elimination</u>

Multiply both equations by an appropriate factor to make the coefficients of one of the variables sum to zero. For example,

7M - 2t = -30   ⇒   -10t + 35M = -150 … (multiply by 5)

5t - 12M = 115   ⇒   10t - 24M = 230 … (multiply by 2)

Now combining the equations eliminates the t terms, and

(-10t + 35M) + (10t - 24M) = -150 + 230

11M = 80

M = 80/11

It follows that

7 × (80/11) - 2t = -30

560/11 - 2t = -30

2t = 890/11

t = 445/11

4 0
2 years ago
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