I. An open-top box can be created by cutting four equal-sized square corners fro
ID: 2886582 • Letter: I
Question
I. An open-top box can be created by cutting four equal-sized square corners from an 8.5 by-inch sheet of paper and folding up the sides (see image below) 8.5I the cutout a. Define a function fthat determines the volume of the box (in cubic inches) provided the length of the side of the square cutout x (in inches). Express the polynomial in both standard and factored form. b. What is an appropriate domain of the function f? c. What is the rate of change of x) with respect to x when the volume of the box is maximized? Explain briefly d. Computer). Use your response to Parts (c) and (d) to determine the dimensions of the box with a maximum volume. e.Explanation / Answer
We are given that the box with an open top is made by cutting equal squares from the corners.
Let each side of the square be = x inches
these square boxes are leter folded to form the cuboidal box with open top.
Hence the dimensions of this box become:
Length = (11-2x) inches
Width = (8.5-2x) inches
Height = x inches
a> Now volume of a cuboid is , V = Length*Width*Height
=> Volume of the box as a function of x is , f(x) = (11-2x)(8.5-2x)x -----> Volume in factored form
or f(x) = 93.5 x - 39 x^2 + 4 x^3 ----------> Volume in standard form
b> The dimensions of the box cannot be zero or negative
=> Length , 11-2x > 0 or x > 11.5/2
=> Width , 8.5 - 2x > 0 or x > 8.5/2
Height, x >0
From the above three inequalities we have the domain of the box as, x > 11.5/2 or x > 5.75
Hence the appropriate domain of the box is , x > 5.75
c> We know the volume of the box is , f(x) = 93.5 x - 39 x^2 + 4 x^3
The rate of change of f(x) is nothig but the derivative of f(x) and for this to maximize we need the below condition :
f '(x) = 0
the reason for the rate of change of f(x) with respect to x to be = 0 is, at maximum value the rate is always = 0
d> we know that f(x) = 93.5 x - 39 x^2 + 4 x^3
=> f '(x) = 93.5 - 78x + 12x2
e> from part we have f '(x) = 93.5 - 78x + 12x2
and at maximum vlome we have f '(x) = 0
=> f '(x) = 93.5 - 78x + 12x2 = 0
=> x = 1.58 and x = 4.91
Next lets find the value of f(x) at the above two points
=> f(x) = 93.5 x - 39 x^2 + 4 x^3
=> f(1.58) = 93.5 (1.58) - 39 (1.58)^2 + 4 (1.58)^3 = 66.15
and f(4.91) = 93.5 (4.91) - 39 (4.91)^2 + 4 (4.91)^3 = -7.65
Hence the volume is maximized at x = 1.58
=> The dimensions that maximize the volume of the box are :
Length = 11.5 - 2x = 11.5 - 2(1.58) = 8.34 inches
Width = 8.5 - 2x = 8.5 - 2(1.58) = 5.34 inches
Height = x = 1.58 inches