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RideAnS [48]
1 year ago
5

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

Mathematics
1 answer:
KiRa [710]1 year ago
7 0

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

<h3>How can the dimensions and volume of the box be calculated?</h3>

The given dimensions of the cardboard are;

Width = 18 inches

Length = 35 inches

Let <em>x </em>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, <em>V</em>, 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, <em>V </em><em> </em>≈ 1048.6 in.³

Learn more about differentiation and integration here:

brainly.com/question/13058734

#SPJ1

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