Respuesta :
Answer:
X= -12 +6√10
X= 6.973
Step-by-step explanation:
Volume of the box= (36-2x)(24-2x)x
Volume of the box
=( 864 -96x -4x²)x
But 864 -96x -4x²
=216 - 24x-x²
Solving for x quadratically
X= (24+12√10)/-2
X= -12 -6√10
X= -30.97
Or
X= (24-12√10)/-2
X= -12 +6√10
X= 6.973
X will definitely be a positive number
So X= -12 +6√10
X= 6.973
Increasing or decreasing the side length of the square that gives the
maximum volume, gives a volume that is less than the maximum.
- The side length of the cut out square to get a box with the maximum volume is approximately 4.71 in.
Reasons:
The given dimensions of the box = 24 in. by 36 in.
Let x represent the dimensions of the square removed from the corners, we have;
Width of box = 24 - 2·x
Length of box = 36 - 2·x
Height of the box = x
Volume of a box = Width × Length × Height
Therefore;
Volume of the box, V = (24 - 2·x)·(36 - 2·x)·x = 4·x³ - 120·x² + 864·x
At the maximum or minimum point of the volume, we have;
- [tex]\displaystyle \frac{dV}{dx} = \mathbf{ \frac{d}{dx} \left(4 \cdot x^3 - 120 \cdot x^2 + 864 \cdot x \right )} = 0[/tex]
Which gives;
12·x² - 240·x + 864 = 0
x² - 20·x + 72 = 0
Which gives;
[tex]\displaystyle x = \dfrac{20\pm \sqrt{(-20)^{2}-4\times 1\times 72}}{2\times 1} = \mathbf{10 \pm 2\cdot \sqrt{7}}[/tex]
At x = 10 + 2·√7, we have;
[tex]\displaystyle 10 + 2 \cdot \sqrt{7} > \frac{24}{2}[/tex]
2 × (10 + 2·√7) in. > 24 in. which is the width of the cardboard
Therefore, (10 + 2·√7) is too long to be cut from the cardboard
At x = 10 - 2·√7, we have;
V = 4·(10 - 2·√7)³ - 120·(10 - 2·√7)² + 864·(10 - 2·√7) ≈ 1,828.3
Therefore;
The maximum volume corresponds with a cut out square of side length x = 10 - 2·√7 inches ≈ 4.71 inches
- The size of the square to be cut out each corner is x ≈ 4.71 inches
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