We have that
<span>points A (-5, 6) and B (7, -1)
Part A)
find the distance
d=</span>√[(y2-y1)²+(x2-x1)²]-------> d=√[(-1-6)²+(7+5)²]----> d=√(49+144)
d=√193 units
Part B)
find the midpoint
ABx=(x1+x2)/2-----> (-5+7)/2-----> ABx=1
ABy=(y1+y2)/2-----> (-1+6)/2-----> ABy=2.5
the midpoint is (1,2.5)
Part C)
find the slope
m=(y2-y1)/(x2-x1)-----> m=(-1-6)/(7+5)--------> m=-7/12
the slope m=-7/12
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:
brainly.com/question/13058734
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