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.
(a) Express the volume V of the box as a function of x
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.) 
(0,−∞)
 
Jul 17th, 2015

Thank you for the opportunity to help you with your question!

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(5-1)(10-1) > 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.

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jul 17th, 2015

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