Consider the following problem: a box with an open top is to be constructed from a square piece of cardboard, 3 ft wide, by cutting out a square from each of the four corners and bending up the sides. find the largest volume that such a box can have. let x denote the length of the side of the square being cut out. let y denote the length of the base.
(a) draw several diagrams to illustrate the situation, some short boxes with large bases and some tall boxes with small bases. find the volumes of several such boxes. (do this on paper. your teacher may ask you to turn in this work.)

(b) draw a diagram illustrating the general situation. introduce notation and label the diagram with your symbols. (do this on paper. your teacher may ask you to turn in this work.)

(c) write an expression for the volume v in terms of x and y.
v =

(d) use the given information to write an equation that relates the variables. (do this on paper. your teacher may ask you to turn in this work.)

(e) use part (d) to write the volume as a function of x.
v(x) =

(f) finish solving the problem by finding the largest volume that such a box can have.
v = ft3

b:  See first attached photo

c:  V = x²y

d and e:  V = x(3 - 2x)²

f:  2 cubic feet

Step-by-step explanation:

a:  Sketch several boxes and calculate the volumes.  

b:  See first attached photo a diagram of this situation

The diagram is a square. We are cutting out squares from the corners.  We don't know the size of the square yet. The side lengths were 3, but now they are 3 - 2x   (since each corner has one side of the square, there are 2 sides of the cut out square on each side of the larger square)

c: The equation for volume is:  V = x²y

The length and width of the box are the x values, the height would be the y value

d and e:  It wants the equation for the volume for our situation.  The base of the box is (3 - 2x)(3 - 2x) or (3 - 2x)².  The height of the box is x, so the volume is

V = x(3 - 2x)²  

f:  Take the derivative, find the critical values, then plug that into x and solve for the volume.  See second attached photo for the work for finding the x value that maximizes the box, and the third attached photo for the evaluation of the maximum volume...

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet