What is the length of a hypotenuse when the legs are 5 and 12
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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
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;
Which gives;
6•x² - 106•x + 315 = 0
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
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