Answer:
x = 3.33... or 10/3
Step-by-step explanation:
3(2x - 4) = 8
distribute the 3 into the parentheses
3 • 2x = 6x
3 • -4 = -12
6x - 12 = 8
add 12 to both sides
6x = 20
divide both sides by 6
x = 3.33... or 10/3
Answer:
The maximum volume of the open box is 24.26 cm³
Step-by-step explanation:
The volume of the box is given as
, where
and
.
Expand the function to obtain:
![g(x)=4x^3-28x^2+48x](https://tex.z-dn.net/?f=g%28x%29%3D4x%5E3-28x%5E2%2B48x)
Differentiate wrt x to obtain:
![g'(x)=12x^2-56x+48](https://tex.z-dn.net/?f=g%27%28x%29%3D12x%5E2-56x%2B48)
To find the point where the maximum value occurs, we solve
![g'(x)=0](https://tex.z-dn.net/?f=g%27%28x%29%3D0)
![\implies 12x^2-56x+48=0](https://tex.z-dn.net/?f=%5Cimplies%2012x%5E2-56x%2B48%3D0)
![\implies x=1.13,x=3.54](https://tex.z-dn.net/?f=%5Cimplies%20x%3D1.13%2Cx%3D3.54)
Discard x=3.54 because it is not within the given domain.
Apply the second derivative test to confirm the maximum critical point.
, ![g''(1.13)=24(1.13)-56=-28.88\:](https://tex.z-dn.net/?f=g%27%27%281.13%29%3D24%281.13%29-56%3D-28.88%5C%3A%3C%5C%3A0)
This means the maximum volume occurs at
.
Substitute
into
to get the maximum volume.
![g(1.13)=1.13(6-2\times1.13)(8-2\times1.13)=24.26](https://tex.z-dn.net/?f=g%281.13%29%3D1.13%286-2%5Ctimes1.13%29%288-2%5Ctimes1.13%29%3D24.26)
The maximum volume of the open box is 24.26 cm³
See attachment for graph.
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Answer: x= 3,2
Step-by-step explanation:
Move all terms to the left side and set equal to zero. Then set each factor equal to zero.