Please help me!! An OPEN BOX PROBLEM??

Question

There's a ton of sections here. Please show me the steps you took to solve these if you can. I want to learn but the way my teacher explains makes these kinds of questions seem impossible.

An open box is to be made from a rectangular piece of cardboard with dimensions 32 inches by 38 inches by cutting out a square of x inches from each corner and turning up the sides. Supply units in each problem where appropriate.

1. Express the volume V of the box as a function of the length x of the side of the square cut from each corner.

2. Express the domain of the function in interval notation. Then, justify this domain as it applies to this specific box problem.

3. Find the length of the side of the square that will be cut out that produces the maximum volume of the box. Then, state the maximum volume obtained. Round your ansers to the nearset thousandths.

4. Give all possible square lengths that could be cut out to yield a volume of 2300 cubic inches. Round your answers to the nearest thousandths.

5. What is the average rate of change of the volume when the length of the square that is cut out changes from 1 inch to 3 inches. Give a clear explanation of how the units correspond in this change.

Answer 1

1)

Write the open box dimensions.

L=38-2x

B=32-2x

H=x

Calculate the volume of the box.

V=LBH

  =(38-2x)(32-2x)(x)

 

2)

V =(38-2x)(32-2x)(x)

   =4x(19-x)(16-x)

Zeros of V(x) are 0,19,16

Possible interval are (0,16)(16,19)(19,∞)

Interval        (0,16)    (16,19)       (19,∞)

Sign of x         +           +               +

Sign of V(x)     +           -               +

Therefore, domain is (0,16)U(19,∞)

Answer 2

3)

For maximum volume, V’ =0

V =4x^3-140x^2+1216x

V’=12x^2-280x+12160

V”=24x-280

x=17.56, 5.77

At x=17.56

V”=0>0

At this value V is minimum.

At x=5.77

V”=0<0

At this value V is maximum.

4)

V=2300

2300=4x^3-140x^2+1216x

4x^3-140x^2+1216x -2300=0

Solve equation.

x=22.75, 9.63, 2.6256

Answer 3

5)

V =4x^3-140x^2+1216x

At x= 1 in

V(1)= 4(1)^3-140(1)^2+1216(1)=1080

At x= 3 in

V(1)= 4(3)^3-140(3)^2+1216(3)=2496

Average rate of change in volume.

AR=[V(3)-V(1)]/(3-1)

     =(2496-1080)/2=708 in^3/s