Can anyone help with te following please
The total volume V(x)cm^3 of a container is given by V(x)= 4(x + �ãx(8-x)) (<0x<8).
Calculate the maximum possible volume of the container?
The shape is two isoseles triangles but connected together to form the container
regards Claire
To calculate the maximum possible volume of the container, we need to find the maximum point of the volume function V(x).
Step 1: Determine the derivative of V(x) with respect to x.
dV(x)/dx = d/dx [4(x + sqrt(x(8-x)))]
To simplify the differentiation, we can rewrite the expression as V(x) = 4x + 4√(8x - x^2).
Using the chain rule, we can differentiate each term separately:
dV(x)/dx = 4 + 4 * d/dx [√(8x - x^2)]
Step 2: Differentiate the square root term.
The derivative of √(8x - x^2) with respect to x can be found by applying the chain rule as follows:
d/dx [√(8x - x^2)] = (1/2) * (8 - 2x) * (d/dx [8x - x^2])
= (1/2) * (8 - 2x) * (8 - 2x)'
= (1/2) * (8 - 2x) * (8-2)
= (1/2) * (8 - 2x) * 6
= 3(8 - 2x)
= 24 - 6x.
Now we have the derivative of V(x) as follows:
dV(x)/dx = 4 + 4 * (24 - 6x)
= 4 + 96 - 24x
= 100 - 24x.
Step 3: Set the derivative equal to zero to find the critical points.
100 - 24x = 0
24x = 100
x = 100/24
x ≈ 4.17 (rounded to two decimal places).
Step 4: Determine if the critical point is a maximum or minimum using the second derivative.
Take the derivative of the derivative calculated in step 2 with respect to x:
d^2V(x)/dx^2 = d/dx (100 - 24x)
= -24.
Since the second derivative is negative, this means the critical point x ≈ 4.17 is a maximum point.
Step 5: Calculate the maximum volume by substituting the critical point into V(x).
V(x) = 4(x + √(x(8-x)))
= 4(4.17 + √(4.17(8-4.17)))
≈ 4(4.17 + √(4.17*3.83))
≈ 4(4.17 + √15.9661)
≈ 4(4.17 + 3.996)
= 4(8.166)
≈ 32.66 cm^3.
Therefore, the maximum possible volume of the container is approximately 32.66 cm^3.