Answer:
a. (30 - 2x)(16 - 2x)x. b. length 21.2 cm , width = 7.2 cm and height x = 4.4 cm
Step-by-step explanation:
a. Let x be the side of the squares to be cut from each corner. Since we have two corners on each side, the length of the resulting box from the 30 cm by 16 cm sheet is L = 30 - 2x. Its breadth is B = 16 - 2x. The height of the resulting box is x. So its volume V = LBx = (30 - 2x)(16 - 2x)x.
b. To find the maximum value of V, we differentiate V with respect to x and equate it to zero.
So, dV/dx = 0
d(30 - 2x)(16 - 2x)x/dx = 0
-2(16 - 2x)x + (-2)(30 - 2x)x + (30 - 2x)(16 - 2x) = 0
32x + 4x² - 60x - 4x² + 480 - 60x - 32x + 4x² = 0
4x²- 120x + 450 = 0
x²- 30x + 112.5 = 0
Using the quadratic formula, we find x. So, with a = 1, b = - 15 and c = 112.5,
[tex]x = \frac{-(-30)+/-\sqrt{(-30)^{2} - 4 X 1 X 112.5} }{2 X 1} \\= \frac{30+/-\sqrt{900 - 450} }{2}\\ = \frac{30+/-\sqrt{450} }{2}\\ = \frac{30+/-21.21 }{2}\\\\ = \frac{30+21.21 }{2} or \frac{30-21.21 }{2}\\ = \frac{51.21 }{2} or \frac{8.79}{2}\\ = 25.605 or 4.395[/tex]
x ≅ 25.61 or 4.4
V = (30 - 2x)(16 - 2x)x
Substituting the values of x into L and B, we have )x
L = (30 - 2x) = (30 - 2(25.61)) = 30 - 51.22 = -21.22
B = (16 - 2x) = (16 - 2(25.61)) = 16 - 51.22 = -35.22
Since L and B cannot be negative, we use the other value for x = 4.4, So
L = (30 - 2x) = (30 - 2(4.4)) = 30 - 8.8 = 21.2
B = (16 - 2x) = (16 - 2(4.4)) = 16 - 8.8 = 7.2
So V = LBx = 21.2 × 7.2 × 4.4 = 671.62 cm³
The dimensions that maximize the volume of the box are 23.34 by 9.34 by 3.33
The dimensions of the material is:
[tex]\mathbf{Length = 30}[/tex]
[tex]\mathbf{Width = 16}[/tex]
Assume the cut-out is x.
So, the dimensions of the box are:
[tex]\mathbf{Length = 30 - 2x}[/tex]
[tex]\mathbf{Width = 16 -2x}[/tex]
[tex]\mathbf{Height =x}[/tex]
The volume is calculated as:
[tex]\mathbf{V = Length \times Width \times Height}[/tex]
So, we have:
[tex]\mathbf{V = (30 - 2x) \times (16 - 2x) \times x}[/tex]
Expand
[tex]\mathbf{V = 480x - 92x^2 +4x^3}\\[/tex]
Differentiate
[tex]\mathbf{V' = 480 - 184x +12x^2}[/tex]
Set to 0
[tex]\mathbf{480 - 184x +12x^2 = 0}[/tex]
Using a calculator, we have:
[tex]\mathbf{x = 12, x =3.33 }[/tex]
12 is greater than the dimensions of the box.
So, we have:
[tex]\mathbf{x =3.33 }[/tex]
Substitute 3.33 for x in
[tex]\mathbf{Length = 30 - 2x}[/tex]
[tex]\mathbf{Width = 16 -2x}[/tex]
[tex]\mathbf{Height =x}[/tex]
So, we have:
[tex]\mathbf{Length = 23.34}[/tex]
[tex]\mathbf{Width = 9.34}[/tex]
[tex]\mathbf{Height = 3.33}[/tex]
Hence, the dimensions that maximize the volume of the box are 23.34 by 9.34 by 3.33
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