Question

Ok Dr. Russ helped me find one part of the question, but I still can't figure out the values of x that maximizes volume = in. Any help is appreciated. Thanks

On a rectangular piece of cardboard with perimeter 11 inches, three parallel and equally spaced creases are made. The cardboard is then folded along the creases to make a rectangular box with open ends. Letting x represent the distance (in inches) between the creases, use a graphing calculator to find the value of that maximizes the volume enclosed by this box. Then give the maximum volume. Round your responses to two decimal places.

values of x that maximizes volume = in
Maximum volume= in^3


I would start with a drawing as I do for most problems. Draw a reactangle 4x one side and b the other.

The perimeter is then 4x+b+4x+b=11

8x+2b=11

if this is folded to a tube then the volume of the tube is bx^2, i.e. a tube with cross sectional area x^2 and length b.

so V=bx^2

rearrangen and substitute for b into the equation above gives

8x+2V/(x^2) = 11

or

8x^3+2V=11x^2

or

V=5.5x^2-4x^3

which you can plot to find max V

I got 1.54 in^3 as the max volume

Answer 1

to find max volume without calculator

dV/dx = 11 x - 12 x^2
= x(11-12 x)
x = 0 or x = 11/12 for max or min (about0.917)

V = 5.5 x^2 -4 x^3 = 1.54 so I agree

Now you had to find x from V instead of doing it the easy way with calculus so
1.54 = 5.5 x^2 - 4 x^3
1.54 = 5.5 x^2 (1 - (4/5.5)x)
1.54 = 5.5 x^2 (1-.727 x)
x^2 (1-.727 x) = .28
now try values of x starting with x = 1
x , left , right
1 , .273 , .28
1.1, .242 , .28 try other direction to get graph
.9 , .280, .28 wow, close !.
.917 , .280 , .28
remember we got .917 with calculus