Question

I'm have some issues with the problem shown in the screenshot and would love some help.

Answer 1

The dimensions and volume of the largest box formed by the 18 in. by 35 in. cardboard are;

• Width ≈ 8.89 in., length ≈ 24.89 in., height ≈ 4.55 in.

• Maximum volume of the box is approximately 1048.6 in.³

How can the dimensions and volume of the box be calculated?

The given dimensions of the cardboard are;

Width = 18 inches

Length = 35 inches

Let x represent the side lengths of the cut squares, we have;

Width of the box formed = 18 - 2•x

Length of the box = 35 - 2•x

Height of the box = x

Volume, V, of the box is therefore;

V = (18 - 2•x) × (35 - 2•x) × x = 4•x³ - 106•x² + 630•x

By differentiation, at the extreme locations, we have;

$\frac{d V }{dx} = \frac{d( 4 \cdot \: {x}^{3} - 106 \cdot \: {x}^{2} + 630\cdot \: {x} ) }{dx} = 0$

Which gives;

$\frac{d V }{dx} =12\cdot \: {x}^{2} - 212\cdot \: {x} + 630 = 0$

6•x² - 106•x + 315 = 0

$x = \frac{ - 6 \pm \sqrt{106 ^2 - 4 \times 6 \times 315} }{2 \times 6}$

Therefore;

x ≈ 4.55, or x ≈ -5.55

When x ≈ 4.55, we have;

V = 4•x³ - 106•x² + 630•x

Which gives;

V ≈ 1048.6

When x ≈ -5.55, we have;

V ≈ -7450.8

The dimensions of the box that gives the maximum volume are therefore;

• Width ≈ 18 - 2×4.55 in. = 8.89 in.

• Length of the box ≈ 35 - 2×4.55 in. = 24.89 in.

• Height = x ≈ 4.55 in.

• The maximum volume of the box, V ≈ 1048.6 in.³

Learn more about differentiation and integration here:

brainly.com/question/13058734

#SPJ1