Best answer:
Starting with a piece of cardboard that is 18" by 22", we have a starting length and width.
If we cut out x" by x" squares from the corners, the length and widths each gets reduced by 2x (since there is a left and right side and a top and bottom side), so the new lengths and widths become:
(18...
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Best answer: Starting with a piece of cardboard that is 18" by 22", we have a starting length and width.
If we cut out x" by x" squares from the corners, the length and widths each gets reduced by 2x (since there is a left and right side and a top and bottom side), so the new lengths and widths become:
(18 - 2x) and (22 - 2x)
with the height of the box being x
The volume in terms of x is then:
V(x) = lwh
V(x) = (18 - 2x)(22 - 2x)x
Simplify and put into standard form:
V(x) = (396 - 36x - 44x + 4x²)x
V(x) = (396 - 80x + 4x²)x
V(x) = (4x² - 80x + 396)x
V(x) = 4x³ - 80x² + 396x
That's your function for volume in terms of x.
For the domain of x, it cannot be zero otherwise you don't have a box.
it can't be half of the smallest dimension since you will no longer have length.
so the domain is:
0 < x < 9
So you can graph the function in this domain using whatever method you know of.
To find the maximum volume, we can solve for the zero of the first derivative and throw out any value of x that is outside of this domain:
V(x) = 4x³ - 80x² + 396x
V'(x) = 12x² - 160x + 396
0 = 12x² - 160x + 396
Divide both sides by 2:
0 = 6x² - 80x + 198
Using quadratic equation:
x = [ -b ± √(b² - 4ac)] / (2a)
x = [ -(-80) ± √((-80)² - 4(6)(198))] / (2 * 6)
x = [ 80 ± √(6400 - 4752)] / 12
x = [ 80 ± √(1648)] / 12
x = [ 80 ± √(16 * 103)] / 12
x = [ 80 ± 4√(103)] / 12
factor out a 4 and cancel out:
x = [ 20 ± √(103)] / 3
Using 10.15 as an approximation for √103, we get:
x = (20 + 10.15) / 3 and (20 - 10.15) / 3
x = 30.15 / 3 and 9.85 / 3
x = 10.05 and 3.283
We said that x had to be less than 9, so we can throw out the one answer, leaving this as the only value:
x = (20 - √103) / 3 inches
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2 days ago