Mathematics, 25.07.2020 20:01 glocurlsprinces

A box with a hinged lid is to be made out of a rectangular piece of cardboard that measures 3 centimeters by 5 centimeters. Six squares will be cut from the cardboard: one square will be cut from each of the corners, and one square will be cut from the middle of each of the -5 centimeter sides . The remaining cardboard will be folded to form the box and its lid . Letting x represent the side-lengths (in centimeters) of the squares, to find the value of that maximizes the volume enclosed by this box. Then give the maximum volume. Round your responses to two decimal places.

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Answer from: myg21

The volume of the box is maximized when x = 0.53 cm

Therefore, x = 0.53 cm and the Maximum volume = 1.75 cm³

Step-by-step explanation:

Please refer to the attached diagram:

The volume of the box is given by

$V = Length \times Width \times Height \\\\$

Let x denotes the length of the sides of the square as shown in the diagram.

The width of shaded region is given by

$Width = 3 - 2x \\\\$

The length of shaded region is given by

$Length = \frac{1}{2} (5 - 3x) \\\\$

So, the volume of the box becomes,

$V = \frac{1}{2} (5 - 3x) \times (3 - 2x) \times x \\\\V = \frac{1}{2} (5 - 3x) \times (3x - 2x^2) \\\\V = \frac{1}{2} (15x -10x^2 -9 x^2 + 6 x^3) \\\\V = \frac{1}{2} (6x^3 -19x^2 + 15x) \\\\$

Take the derivative of volume and set it to zero.

$\frac{dV}{dx} = 0 \\\\\frac{dV}{dx} = \frac{d}{dx} ( \frac{1}{2} (6x^3 -19x^2 + 15x)) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\0 = \frac{1}{2} (18x^2 -38x + 15)\\\\18x^2 -38x + 15 = 0\\\\$

We may solve the quadratic equation using the quadratic formula.

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

The values of coefficients a, b, c are

$a = 18 \\\\b = -38 \\\\c = 15 \\\\$

Substituting the values into quadratic formula yields,

$x=\frac{-(-38)\pm\sqrt{(-38)^2-4(18)(15)}}{2(18)} \\\\x=\frac{38\pm\sqrt{(1444- 1080}}{36} \\\\x=\frac{38\pm\sqrt{(364}}{36} \\\\x=\frac{38\pm 19.078}{36} \\\\x=\frac{38 + 19.078}{36} \: or \: x=\frac{38 - 19.078}{36}\\\\x= 1.59 \: or \: x = 0.53 \\\\$

Volume of box when x = 1.59:

$V = \frac{1}{2} (5 - 3(1.59)) \times (3 - 2(1.59)) \times (1.59) \\\\V = -0.03 \: cm^3 \\\\$

Volume of box when x = 0.53:

$V = \frac{1}{2} (5 - 3(0.53)) \times (3 - 2(0.53)) \times (0.53) \\\\V = 1.75 \: cm^3$

As you can see, the volume of the box is maximized when x = 0.53 cm

Therefore, x = 0.53 cm and the Maximum volume = 1.75 cm³

A box with a hinged lid is to be made out of a rectangular piece of cardboard that measures 3 centim

Answer from: Quest

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

Answer from: Quest

the answer is d

the answer is in the first quadrant, which make the y-int positive.

since the dots are going down, therefore the correlation is negative