The volume of an open-top box is given by V(x)= 8x(6-x)(12-x)
a) Give the domain for x.
b) Find the x that maximizes the volume for the box?
c) What is the maximum volume in cubic inches?
See:
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a cylinder has the volume of 224 pie cubic inches and a radius of 4 inches whats the height
To answer the given questions, we need to analyze the volume function V(x) = 8x(6-x)(12-x).
a) Domain for x:
The domain of x represents the set of values that x can take in the given volume function. In this case, since V(x) represents the volume of a physical object, x must have real and positive values. Additionally, since the box has an open-top, the height of the box cannot be greater than 12 inches. This corresponds to x ≤ 12. Therefore, the domain for x is x ≥ 0 and x ≤ 12.
b) Finding the x that maximizes the volume for the box:
To find the x that maximizes the volume, we need to find the critical points of the function by taking its derivative and setting it equal to zero. Let's calculate the derivative of V(x):
V'(x) = 8(6-x)(12-x) + 8x(-1)(12-x) + 8x(6-x)(-1)
Expanding and simplifying further, we have:
V'(x) = (72 - 8x - 12x + x²) + (-8x + 8x²) + (-8x + x²)
= 72 + x² - 20x - 16x + 8x² - 8x + x²
= 10x² - 44x + 72
To find the critical points, we set V'(x) equal to zero and solve for x:
10x² - 44x + 72 = 0
This is a quadratic equation, and we can solve it by factoring, completing the square, or using the quadratic formula. After solving the equation, we will obtain two values of x. These x-values correspond to potential critical points.
c) Finding the maximum volume:
Once we have the critical points, we need to evaluate the volume function V(x) at each critical point, as well as at the endpoints of the domain (x = 0 and x = 12). By comparing the values obtained, we can determine the x-value that yields the maximum volume. Finally, we substitute this x-value back into the volume function to find the maximum volume in cubic inches.