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

*label*Algebra

*account_circle*Unassigned

*schedule*1 Day

*account_balance_wallet*$5

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*of the box as a function of

*x*.

V(x) = | 4 x(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.)

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.

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