Solve Two Step Eqautions

Since the dimensions of the square piece of paper are 18 inches each, and we cut out a length x from each side to give a total length of 2x cut from each side. So, each dimension is L = 18 - 2x.

Since the height of the open box is x and its base is a square, the volume of the open box ix V = L²x

= (18 - 2x)²x

= (324 - 72x + 4x²)x

= 324x - 72x² + 4x³

The volume of the box V = 4x³ - 72x² + 324x

b.

The minimum value of length of the sides of the squares cut from each corner is x = 3 inches.

the length of the box, L = 12 inches,
the width of the box = 12 inches
and the height of the box, x = 3 inches.

Since the volume of the box is 432 cubic inches,

V = 324x - 72x² + 4x³

324x - 72x² + 4x³ = 432

4x³ - 72x² + 324x - 432 = 0

x³ - 18x² + 81x - 108 = 0

A factor of the expression is x - 3

So, x³ - 18x² + 81x - 108 ÷ x - 3 = x² - 15x + 36

So, x³ - 18x² + 81x - 108 = (x² - 15x + 36)(x - 3) = 0

Factorizing the expression x² - 15x + 36 = 0

x² - 3x - 12x + 36 = 0

x(x - 3) - 12(x - 3) = 0

(x - 3)(x - 12) = 0

So, x³ - 18x² + 81x - 108 = (x - 3)(x - 3)(x - 12) = 0

So, (x - 3)²(x - 12) = 0

(x - 3)² = 0 and (x - 12) = 0

x - 3 = √0 and x - 12 = 0

x - 3 = 0 and x - 12 = 0

x = 3 twice and x = 12

Since x = 3 is the minimum value, the minimum value of x = 3.

Since the length of the box, L = 18 - 2x

= 18 - 2(3)

= 18 - 6

= 12 inches

The width of the box = L = 12 inches

The height of the box = x = 3 inches.

So,

The minimum value of length of the sides of the squares cut from each corner is x = 3 inches.

the length of the box, L = 12 inches,
the width of the box = 12 inches
and the height of the box, x = 3 inches.

Learn more about volume of a box here:

brainly.com/question/13309609

6 0

2 years ago

Find an equation of the sphere containing all surface points P = (x, y, z) such that the distance from P to A(−3, 6, 3) is "twic

Marina86 [1]

Answer:

Equation of Sphere = $x^{2}$ + $y^{2}$ + $z^{2}$ - 18x - 4/3y +10z + 142/3 = 0

Step-by-step explanation:

Data Given:

P = (x,y,z)

Distance from P to A (-3,6,3) = Twice the distance from P to B(6,2,-3)

Solution:

Find the equation of the sphere:

It is given that:

PA = 2PB

Squaring both sides:

$(PA)^{2} = (2PB)^{2}$

$(PA)^{2} = 4 (PB)^{2}$

$(x - (-3))^{2}$ + $(y - 6)^{2}$ + $(z-3)^{2}$ = 4 x $(x-6)^{2}$ + $(y-2)^{2}$ + $(z-(-3))^{2}$

Solving the above equation:

$(x + 3))^{2}$ + $(y - 6)^{2}$ + $(z-3)^{2}$ = 4 x { $(x-6)^{2}$ + $(y-2)^{2}$ + $(z + 3))^{2}$ }

$x^{2}$ + 9 + 6x + $y^{2}$ + 36 - 12y + $z^{2}$ + 9 - 6z = 4 { $x^{2}$ + 36 - 12x + $y^{2}$ + 4 - 4y + $z^{2}$ + 9 + 6z}

$x^{2}$ + 9 + 6x + $y^{2}$ + 36 - 12y + $z^{2}$ + 9 - 6z = 4 $x^{2}$ + 144 - 48x +4 $y^{2}$ + 16 - 16y + 4 $z^{2}$ + 36 + 24z

Putting the right hand side = 0 and solving the equation:

$x^{2}$ - 4 $x^{2}$ + $y^{2}$ - 4 $y^{2}$ + $z^{2}$ - 4 $z^{2}$ + 6x + 48x - 12y +16y -6z - 24z + 9 + 36 + 9 -144 - 16 - 36 = 0

-3 $x^{2}$ - 3 $y^{2}$ -3 $z^{2}$ + 54x + 4y -30z -142 = 0

Taking (-) common

- ( 3 $x^{2}$ + 3 $y^{2}$ + 3 $z^{2}$ - 54x - 4y +30z + 142) = 0

3 $x^{2}$ + 3 $y^{2}$ + 3 $z^{2}$ - 54x - 4y +30z + 142 = 0

dividing the whole equation by 3

$x^{2}$ + $y^{2}$ + $z^{2}$ - 54/3x - 4/3y +30/3z + 142/3 = 0

$x^{2}$ + $y^{2}$ + $z^{2}$ - 18x - 4/3y +10z + 142/3 = 0

7 0

3 years ago