Can anyone help with te following please
The total volume V(x)cm^3of a container is given by V(x)=4(x+�ãx(8-x)) (<0x<8).
Calculate the maximum possible volume of the container?
regards Claire
�ãx(8-x) ?
Please edit your problem.
V(x)=4(x+√x(8-x)) (<0x<8)
To calculate the maximum volume of the container, we need to find the maximum value of the given function V(x).
The given function is V(x) = 4(x + √(x(8-x))) for 0 < x < 8.
To find the maximum value, we can use calculus. We need to find the derivative of V(x) with respect to x and then set it equal to zero to find the critical points.
1. First, let's find the derivative of V(x):
V'(x) = 4(1 + [(8-x) - x]) / 2√(x(8-x))
= 4(1 - x) / 2√(x(8-x))
= 2 (1 - x) / √(x(8-x))
2. Next, we set V'(x) equal to zero and solve for x:
2 (1 - x) / √(x(8-x)) = 0
Since the denominator √(x(8-x)) cannot be zero, we can conclude that x = 1 satisfies this equation.
3. Now, we need to determine if this critical point is a maximum or minimum. We can use the second derivative test for this purpose. Let's find V''(x):
V''(x) = d/dx [2 (1 - x) / √(x(8-x))]
= 2 [0 - (-1) / 2√(x(8-x))^3
= (1 - x) / √(x(8-x))^3
4. Now, substitute the critical point x = 1 into V''(x):
V''(1) = (1 - 1) / √(1(8-1))^3
= 0
Since V''(1) = 0, the second derivative test fails to determine if the critical point x = 1 is a maximum or minimum.
5. Therefore, the last step is to check the endpoints of the given interval, which are x = 0 and x = 8.
When x = 0, V(0) = 4(0 + √(0(8-0)))
= 4(0 + √(0))
= 4(0)
= 0
When x = 8, V(8) = 4(8 + √(8(8-8)))
= 4(8 + √(8(0)))
= 4(8 + √(0))
= 4(8 + 0)
= 4(8)
= 32
Comparing the volume at x = 0 and x = 8, we can see that V(0) = 0 and V(8) = 32.
6. Now, we compare these values to the value of the critical point x = 1. Since V(1) is not equal to 0 or 32, it means that V(1) represents the maximum volume of the container.
Therefore, the maximum possible volume of the container is V(1) = 4(1 + √(1(8-1))) = 4(1 + √(1(7))) = 4(1 + √(7)) = 4(1 + 2.6457) = 4(3.6457) = 14.5828 cm^3.
So, the maximum possible volume of the container is approximately 14.5828 cm^3.