SOLUTION: On a rectangular piece of cardboard with perimeter 10 inches, three parallel and equally spaced creases are made. The cardboard is then folded along the creases to make a rectangul
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Question 1149561: On a rectangular piece of cardboard with perimeter 10 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 x that maximizes the volume enclosed by this box. Then give the maximum volume. Round your responses to two decimal places.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
On a rectangular piece of cardboard with perimeter 10 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 x that maximizes the volume enclosed by this box.
Then give the maximum volume. Round your responses to two decimal places.
:
let L = the length of the cardboard
let x = dist between creases
then
2L + 8x = 10
simplify divide by 2
L + 4x = 5
L = (5-4x)
:
Volume
V = x^2 * L
Replace L with (5-4x)
V = x^2(5-4x)
V = -4x^3 + 5x^2
:
Graph this equation
Max volume when x = .83 inches
Find the volume
V = -4(.83^3) + 5(.83^2)
V = -2.29 + 3.45
V = 1.16 cu/in is the max volume (green line)
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