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Murljashka [212]
3 years ago
11

| 3 x - 2 | = 4x + 4

Mathematics
1 answer:
Ugo [173]3 years ago
5 0

Answer: -2/7

|3x - 2| - 4x = 4

1) (3х - 2) - 4х = 4, if 3x - 2 >= 0

2) -(3x - 2) - 4x = 4, if 3x - 2 < 0

1)

3х - 4х = 4 + 2

-x = 6

x = -6

3х - 2 >= 0

3х >= 2

x >= 2/3 - wrong

2)

-3х + 2 - 4х = 4

-7х = 2

x = -2/7

3x-2<0

3x<2

3(-2/7)<2-right

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Answer:

The dimension of the box is 17.66 in by 7.66 in by 2.67 in.

Therefore the volume of the box is 361.19 in^3.

Step-by-step explanation:

Given that the dimensions of a cardboard is 23 in by 13 in.

Let the side of the square be x in.

Then the length of the box= (23-2x) in

and the width of the box =(13-2x) in

and height = x in.

The volume of the box is = length ×width × height

                                         =[(23-2x)(13-2x)x] in^3

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∴V=299x-72x² +4x³

Differentiating with respect x

V'= 299-144x+12x²

Again differentiating with respect x

V''= -144+24x

To find the dimensions, we set V'=0

∴299-144x+12x²=0

Applying Sridharacharya formula that is the solution of a quadratic equation ax²+bx+c is x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

Here a=12, b=-144 , c=299

\therefore x=\frac{-(-144)\pm\sqrt{(-144)^2-4.12.299}}{2.12}

\Rightarrow x= 9.33, 2.67

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∴x = 2.67 in

If at x = 2.67, V''<0 , then the volume of the box will be maximum or V''>0 then volume of the box will be minimum.

V''|_{x=2.67}=-144+(24\times 2.67)=-79.92

Therefore  at x = 2.67, the volume of the box maximum.

The length of the box =[23-(2×2.67)] in

                                    =17.66 in

The width of the box =[13-(2×2.67)] in

                                  =7.66 in

The height of the box= 2.67 in

The dimension of the box is 17.66 in by 7.66 in by 2.67 in.

Therefore the volume of the box is =(17.66×7.66×2.67) in^3

                                                            =361.19 in^3

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