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

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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) =  4x(10−x)(5−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 dimensions of the box would be:
length = (20  2x) = 2(10  x)
width = (10  2x) = 2(5  x)
height = x
Then the volume would be:
V = (height)(width)(length) = (x)*2(5  x)*2(10  x) > V = 4x(5  x)(10  x)
So part a would be: V = 4x(5  x)(10  x)
For part b our volume must be greater than zero, so we have:
4x(5  x)(10  x) > 0
So we equal each factor to zero and solve for x like this.
4x = 0 > 4x/4 = 0/4 > x = 0
5  x = 0 > 5  x + x = 0 + x > 5 = x > x = 5
10  x = 0 > 10  x + x = 0 + x > 10 = x > x = 10
Then we graph it on the real line so we can see how many intervals we have and we can choose a test point for each interval.
infinity  x=0  x=5  x=10  Infinity
So from infinity to 0 we can choose x = 1.
From x = 0 to x = 5 we can choose x = 1.
From 5 to 10 we choose x = 6.
From 10 to infinity we can choose x = 11.
Then we enter each test point into the original inequality and if it is satisfied (it's true) then it is a part of the solution. So we would have:
For x = 1
4x(5  x)(10  x) > 0 > 4(1)(5 (1))(10  (1)) > 0 > 4(5+1)(10+1) > 0 > 264 > 0. False (this interval is not a part of the solution).
For x = 1
4x(5  x)(10  x) > 0 > 4(1)(5 1)(10 1) > 0 > 4(51)(101) > 0 > 144 > 0. True (the interval from 0 to 5 is a part of the solution).
For x = 6
4x(5  x)(10  x) > 0 > 4(6)(5 6)(10 6) > 0 > 24(1)(4) > 0 > 96 > 0. False.
For x = 11
4x(5  x)(10  x) > 0 > 4(11)(5 11)(10 11) > 0 > 44(6)(1) > 0 > 264 > 0. True (the interval from 10 to infinity is a part of the solution).
Then we would have: (0 , 5) since the length must be positive.
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