Mathematics, 15.04.2020 03:35 kiyomi58

. Squares with sides of length x are cut out of each corner of a rectangular piece of cardboard measuring 2323 ft by 1313 ft. The resulting piece of cardboard is then folded into a box without a lid. Find the volume of the largest box that can be formed in this way. b. Suppose that in part (a) the original piece of cardboard is a square with sides of length s. Find the volume of the largest box that can be formed in this way.

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Answer from: andesmints5341

The volume of a box is the amount of space in it.

The maximum value of the box is $\mathbf{361.186in^3}$ The volume of the largest box is $\mathbf{22.559in^3}$

The dimension of the cardboard is given as:

$\mathbf{Length = 23}$

$\mathbf{Width = 13}$

Assume the cut-out is x.

So, the dimension of the box is:

$\mathbf{Length = 23 -2x}$

$\mathbf{Width = 13 - 2x}$

$\mathbf{Height = x}$

The volume of the box is:

$\mathbf{V = (23 -2x) (13 - 2x)x}$

Expand

$\mathbf{V = 299x -72x^2 +4x^3}$

Differentiate

$\mathbf{V' = 299 -144x +12x^2}$

Set to 0

$\mathbf{299 -144x +12x^2 = 0}$

Rewrite as:

$\mathbf{12x^2 - 144x + 299= 0}$

Using a calculator, we have:

$\mathbf{x = (2.67, 9.33)}$

9.33 is greater than the dimensions of the box.

So, the possible value of x is:

$\mathbf{x = 2.67}$

Recall that:

$\mathbf{V = 299x -72x^2 +4x^3}$

So, we have:

$\mathbf{V = 299(2.67) - 72(2.67)^2 + 4(2.67)^3}$

$\mathbf{V = 361.186}$

Hence, the maximum value of the box is $\mathbf{361.186in^3}$

If the dimension of the cardboard is s,

The dimension of the box would be:

$\mathbf{Length = s -2x}$

$\mathbf{Width = s - 2x}$

$\mathbf{Height = s}$

So, the volume is:

$\mathbf{V = (s - 2x)(s - 2x) s}$

Substitute $\mathbf{x = 2.67}$

$\mathbf{V = (s - 2 \times 2.67)(s - 2 \times 2.67) s}$

$\mathbf{V = (s - 5.34)(s - 5.34) s}$

Expand

$\mathbf{V = (s - 5.34)(s^2 - 5.34s)}$

$\mathbf{V = s^3 - 5.34s^2 - 5.34s^2 + 28.5156s}$

$\mathbf{V = s^3 -10.68s^2 + 28.5156s}$

Integrate

$\mathbf{V' = 3s^2 -21.36s + 28.5156}$

Set to 0

$\mathbf{3s^2 -21.36s + 28.5156 = 0}$

Using a calculator, we have:

$\mathbf{s = (1,78,5.34)}$

5.34 is greater than the side lengths. So, we have:

$\mathbf{s = 1.78}$

Recall that:

$\mathbf{V = s^3 -10.68s^2 + 28.5156s}$

So, we have:

$\mathbf{V = 1.78^3 - 10.68 \times 1.78^2 + 28.5156 \times 1.78}$

$\mathbf{V = 22.559}$

Hence, the volume of the largest box is $\mathbf{22.559in^3}$

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