Question

3) Postcards: Marcie wants to make a box to hold her postcard collection from a piece of

Answer 1

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

The dimension of the postcard is:

$\mathbf{Length = 10}$

$\mathbf{Width = 18}$

Let x represents the length of the cut.

So, the dimension of the box is:

$\mathbf{Length = 10 -2x}$

$\mathbf{Width = 18 -2x}$

$\mathbf{Height = x}$

(a) The function that represents the volume of the box

This is calculated as:

$\mathbf{V(x) = Length \times Width \times Height}$

So, we have:

$\mathbf{V(x) = (10 - 2x) \times (18 - 2x) \times x}$

Expand

$\mathbf{V(x) = 180x - 56x^2 + 4x^3}$

Hence, the volume function is: $\mathbf{V(x) = 180x - 56x^2 + 4x^3}$

(b) The dimension that maximizes the volume

We have:

$\mathbf{V(x) = 180x - 56x^2 + 4x^3}$

Differentiate

$\mathbf{V'(x) = 180- 112x + 12x^2}$

Set to 0

$\mathbf{180- 112x + 12x^2 = 0}$

Using a calculator, we have:

$\mathbf{x = 7.3,2.1}$

The value x = 7.3 is greater than the dimension of the box.

So, we have:

$\mathbf{x = 2.1}$

Recall that:

$\mathbf{Length = 10 -2x}$

$\mathbf{Width = 18 -2x}$

$\mathbf{Height = x}$

So, we have:

$\mathbf{Length = 10 -2 \times 2.1 = 5.8}$

$\mathbf{Width = 18 -2 \times 2.1 = 13.8}$

$\mathbf{Height = 2.1}$

Hence, the dimension that maximizes the volume is 5.8 by 13.8 by 2.1

(c) The maximum volume

This is calculated as the product of the dimension of the box

So, we have:

$\mathbf{Volume = 5.8 \times 13.8 \times 2.1}$

$\mathbf{Volume = 168.1}$

Hence, the maximum volume of the box is 168.1 cubic inches

Read more about volumes at:

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