A graphing calculator is recommended. An open box is to be constructed from a
Algebra

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A graphing calculator is recommended.
An open box is to be constructed from a piece of cardboard 10 cm by 20 cmlength x from each corner and folding up the sides, as shown in the figure.
V(x) = 
(b) What is the domain of V? (Use the fact that length and volume must be positive. Enter your answer using interval notation.)
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The dimension of the cube would be: (10  2x) , (20  2x) , and x (see the picture). Then we fin the volumen of the cube by using this formula:
V = (length)(width)(height) > V (x) = (20  2x)(10  2x)x
We could factor 2 from the expressions 20  2x and 10  2x like this:
V(x) = 2(10  x)*2(5  x)x > V(x) = 4x(10  x)(5  x). This is part A.
Now for part B first we need to equal each factor to zero and solve for x like this:
4x(10  x)(5  x) = 0 > 4x = 0 , 10  x = 0 , 5  x = 0 . So we solve each equation for x.
4x = 0 > 4x/4 = 0/4 > x = 0
10  x = 0 > 10  x + x = 0 + x > 10 = x > x = 10
5  x = 0 > 5  x + x = 0 + x > 5 = x > x = 5
Then we graph the obtained x values on the real line to see how many intervals we have.
 inf  x= 0  x =5  x =10  infinity
So as we can see we have 4 intervals which are: 1 st interval from  infinity to 0, the 2nd interval from 0 to 5, the 3rd interval from 5 to 10, and the last interval from 10 to infinity. So we choose a test point for each interval (e.g. for the 1st interval x = 1 since it is between  infinity and 0, and so on) and we enter it into the inequality and if it is satisfied then it belongs to the solution.
4x(10  x)(5  x) > 0 (since the volume must be greater than zero).
For x = 1
4(1)(10 (1))(5  (1)) > 0 > 4(11)(6) > 0 > 264 > 0 False
For x = 1
4(1)(10 1)(5 1) > 0 > 4(9)(4) > 0 > 144 > 0 True
For x = 2 False
For x = 11 True
Then the domain would be: (0, 5)U(
It is missing the last part of the answer. The domain is:
(0 , 5) U (10, infinity)
Please let me know if you have any doubt or question.
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