HELP!
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?
V(x)= 8x(6-x)(12-x)
The domain contains all permissible values of the function in the particular application. Since the box cannot have negative sides, so the domain should reflect that, nor can the side be negative.
Thus the domain of x is limited to values between 0 and 6, or (0,6), since a box of zero volume is meaningless.
Expand V(x) to a polynomial and differentiate to get the maximum volume. Do not omit to check that the value is indeed a maximum and not a minimum.
The polynomial, when expanded, is
8x^3-144x^2+576x
a) To determine the domain for x, we need to identify any restrictions on the values of x. In this case, the volume function is given by V(x) = 8x(6-x)(12-x).
Since volume is a physical quantity, it must be non-negative. Therefore, the domain for x is the set of values for which the volume function is non-negative.
To find the domain, we set V(x) >= 0 and solve for x:
8x(6-x)(12-x) >= 0
We can use a sign chart or test intervals to find the values of x for which the inequality holds.
First, let's identify the critical points by setting each factor equal to zero:
x = 0, 6, and 12
Next, we choose test points within the intervals determined by the critical points:
For the interval (-∞, 0), we can choose x = -1.
For the interval (0, 6), we can choose x = 3.
For the interval (6, 12), we can choose x = 9.
For the interval (12, +∞), we can choose x = 13.
Evaluating V(x) at these test points:
For x = -1: V(-1) = 8(-1)(6-(-1))(12-(-1)) = -560, which is negative.
For x = 3: V(3) = 8(3)(6-3)(12-3) = 432, which is positive.
For x = 9: V(9) = 8(9)(6-9)(12-9) = -432, which is negative.
For x = 13: V(13) = 8(13)(6-13)(12-13) = 560, which is positive.
From the above evaluations, we observe that V(x) is positive in the intervals (0, 6) and (12, +∞), and negative in the intervals (-∞, 0) and (6, 12). Therefore, the domain for x is:
Domain: (0, 6) U (12, +∞)
b) To find the value of x that maximizes the volume of the box, we need to find the critical points where the derivative of V(x) is equal to zero or undefined.
Let's start by finding the derivative of V(x):
V'(x) = 8[(6-x)(12-x) + x(-1)(12-x) + x(6-x)(-1)].
Now, set V'(x) = 0 and solve for x:
8[(6-x)(12-x) + x(-1)(12-x) + x(6-x)(-1)] = 0.
Simplifying the equation:
(6-x)(12-x) + x(-1)(12-x) + x(6-x)(-1) = 0.
Expanding and rearranging:
(6-x)(12-x) - (12-x)(x) - (6-x)(x) = 0.
Simplifying further:
(6-x)(12-x - x) = 0.
(6-x)(12-2x) = 0.
The critical points are obtained by setting each factor equal to zero:
6-x = 0, which gives x = 6.
12-2x = 0, which gives x = 6.
Therefore, the critical point is x = 6.
c) To determine the maximum volume, we need to evaluate the volume function at the critical points and the endpoints of the domain.
V(x) = 8x(6-x)(12-x).
V(0) = 8(0)(6-0)(12-0) = 0.
V(6) = 8(6)(6-6)(12-6) = 0.
V(12) = 8(12)(6-12)(12-12) = 0.
We observe that the volume is zero at x = 0, x = 6, and x = 12.
Next, let's find the values at the endpoints of the domain:
V(0+) = 8(0+)(6-0+)(12-0+) = 0.
V(∞) = lim(x->∞) 8x(6-x)(12-x) = -∞.
We can see that the volume approaches negative infinity as x approaches infinity.
Therefore, the maximum volume occurs at one of the critical points, x = 6.
The maximum volume is given by V(6):
V(6) = 8(6)(6-6)(12-6) = 0.
Hence, the maximum volume of the box is 0 cubic inches.
a) To determine the domain for x, we will look at the given expression V(x) = 8x(6-x)(12-x). The volume of a box cannot be negative, so we need to ensure that the expression inside the volume function is greater than or equal to zero.
Let's solve for the factors: (6-x), (12-x).
For (6-x) to be greater than or equal to zero, we have: 6-x >= 0.
Solving for x, we find x <= 6.
For (12-x) to be greater than or equal to zero, we have: 12-x >= 0.
Solving for x, we find x <= 12.
Taking the intersection of these two conditions, we find that the domain for x is 0 <= x <= 6.
b) To find the value of x that maximizes the volume of the box, we need to find the critical points. Critical points occur where the derivative of the function is equal to zero or undefined.
Firstly, let's simplify the expression for V(x): V(x) = 8x(6-x)(12-x).
To find the critical points, we need to differentiate the function with respect to x. So, let's differentiate V(x) using the product rule:
V'(x) = 8(6-x)(12-x) + 8x(-1)(12-x) + 8x(6-x)(-1).
Simplifying this expression, we have: V'(x) = 8(6-x)(12-x) - 8x(12-x) - 8x(6-x).
Expanding and simplifying further, we get: V'(x) = 96 - 20x + 2x^2 - 8x^2.
Combining like terms, we get: V'(x) = 96 - 28x - 6x^2.
To find the critical points, we set V'(x) equal to zero and solve for x:
0 = 96 - 28x - 6x^2.
This equation is quadratic, so let's solve it by factoring or using the quadratic formula.
Alternatively, we can use a graphing calculator or software to find the x-coordinate of the maximum point on the graph of V(x).
c) Once we find the value of x that maximizes the volume of the box, we substitute this value back into the volume function V(x) = 8x(6-x)(12-x) to calculate the maximum volume in cubic inches.