This was answered by DrRuss on 1/21 at 6:39 am. I can't link so I pasted his answer below.
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
but check the maths!
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'm so lost and don't know where to start any help would be greatly appreciated
3 answers
Ok but I still don't know how to get the values of x that maximizes volume = in
sorry, I didn't know that.
you should have said that in your 1st post above. the tutors would see that this question was answered and refer you to the original answer.
you should have said that in your 1st post above. the tutors would see that this question was answered and refer you to the original answer.