Answer:
0.785 m
Step-by-step explanation:
(4-2x)(6-2x)x = Volume
Volume = 4x³ - 20x² + 24x
Max volume: dV/dx = 0
dV/dx = 12x² - 40x + 24 = 0
3x² - 8x + 6 = 0
x = 2.54858377, 0.784749563
Max when d²V/dx² < 0
d²V/dx² = 6x - 8
At x = 0.784749563
6x - 8 = -3.29 < 0
Given that you are constructing a cardboard box with the dimensions 4 m by 6 m. You then cut equal-size squares (x by x squares) from each corner so that you may fold the edges to get a box. x should be 0.785 m to get a box with the maximum volume of cuboid.
The volume of a cuboid is found by multiplying the length by the breadth by the height.
Given the cardboard box with the dimensions 4 m by 6 m and we have cut x by x squares from each corner to form cuboid.
Dimensions of the cuboid =
Length = (6 - 2x) m
Breadth = (4 - 2x) m
Height = x m
Volume of the cuboid = x(6 - 2x)(4 - 2x) m³
To maximize the volume, we set it equal to a function V.
V = x(6 - 2x)(4 - 2x) = 4x(3 - x)(2 - x) = 24x - 20x² + 4x³
Put the first derivative equal to 0.
24 - 40x + 12x² = 0
6 - 10x + 3x² = 0
3x² - 10x + 6 = 0
Solving the quadratic equation to find x.
[tex]x = \frac{-b +- \sqrt{b^{2}-4ac } }{2a}[/tex]
[tex]x = \frac{10 +- \sqrt{100 - 72} }{6} \\\\x = \frac{10 +- 2\sqrt{7} }{6} \\\\x = \frac{5 + \sqrt{7} }{3} , \frac{5 - \sqrt{7} }{3}[/tex]
x = 2.549, 0.785
We use the second derivative test.
Second derivative of V = 24 - 40x + 12x²
Putting the values of x in here.
V"(2.549) = 0.008812
V"(0.785) = -0.0053
Since V"(0.785) < 0, maximum volume exist at x = 0.785.
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