A rectangular piece of tin has an area of 1334 square inches. A square tab of 3 inches is cut from

Question

each corner, and the ends and sides are turned up to make an open box. If the volume of the box is
2760 cubic inches, what were the original dimensions of the rectangular piece of tin? Show the work
that leads to the answer.
2. Now, let’s maximize the volume of the box that can be made using the original dimensions you found
in problem #1. Carefully complete the sequence below.
a) Find the algebraic representation for the volume of the box, V(x), where x is the length of the square tab
in each corner. Express V(x) as a polynomial in factored form and standard form. Make a table of values
for x and V if the side length of the square tab is x = 1, 3, 5, 7, 9, 11, and 13 inches. Show the work that
leads to each answer. How does the volume of the box change when the size of the square tabs increases
in size?
b) Length, width, and height must be positive. Use this fact to find the domain of the volume function V(x)
in the form a < x < b. Explain. Justify your explanation by solving inequalities.
c) Find the algebraic representation for the area of the bottom of the box, A(x). Express A(x) as a
polynomial in factored form and standard form. Make a table of values for x and A if the side length of
the square tab is x =1, 3, 5, 7, 9, 11, and 13 inches. Show the work that leads to each answer. How does
the area of the bottom of the box change when the size of the square tabs increases in size?
d) Use a graphing calculator to sketch the graph of y = V(x). Use the viewing window [0, 24, 4] by
[-2500, 4000, 500]. Then use the 2
nd
– TRACE – maximum feature to calculate its maximum point.
What are the dimensions of the box, (l  w  h), with the maximum volume? What is the maximum
volume of the box? Draw and label a graph to support this answer. Label the coordinates of the x-and yintercepts
and relative maximum and minimum points. Values must be accurate to three decimal places.
e) Find and simplify the difference quotient of y = V(x). That is, find ௏(௫ା௛)ି௏(௫)
௛
. (This will get a little
messy.) Then, substitute h with 0 and simplify. The expression you obtained is called the derivative of
y = V(x). (You will study derivatives in calculus.) Now, find the zeros of the derivative accurate to three
decimal places. What do you notice?

Answer 1

Wow, nice problem with very clear and precise instructions how to do it.
I suggest you just follow the steps they ask you to perform.

I will start you off.

Let the finished box be x inches long, y inches wide and 3 inches high

given: 3xy = 2760
xy = 920

original piece is (x+6) by (y+6)
(x+6)(y+6) = 1334
xy + 6x + 6y + 36 = 1334
920 + 6(x+y) = 1298
6(x+y) = 378
x+y = 63 or y = 63-x

sub into xy= 920
x(63-x) = 920
63x - x^2 - 920 = 0
x^2 - 63x + 920 = 0
(x-40)(x-23) = 0

x = 40 inches or x = 23 inches
If x=40, then y = 23
If x = 23, then y = 40

so the rectangle was 46 by 29 inches

2. now let the tab to be cut out be x inches
(this x is not the same as the x we just used)

etc. carry on

Answer 2

Thanks for the help, but do you understand what Im suppoosed to graph on 2D d) Use a graphing calculator to sketch the graph of y = V(x). Use the viewing window [0, 24, 4] by
[-2500, 4000, 500]. Then use the 2
nd
– TRACE – maximum feature to calculate its maximum point.
What are the dimensions of the box, (l  w  h), with the maximum volume? What is the maximum