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;
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.
- The maximum volume of the box, <em>V </em><em> </em>≈ 1048.6 in.³
Learn more about differentiation and integration here:
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Answer:
20
Step-by-step explanation:
because 4 can go into 20 5 times without a remainder
<h3>
Answer: Point U is between points N and S.</h3>
The order S, U, N shows U is in the middle. We can read this in reverse to get N, U, S.
Notice how the letter "U"s are adjacent in NU and US to get NU+US.
Answer:
(-3+) /2, (3+) / 2 which is approx -3.791 and 0.791
Step-by-step explanation:
i used the quadratic formula: [-b± sq rt (b² - 4ac) / 2a]
in this problem, a = 1, b = 3, and c = -3