SOLUTION: On a rectangular piece of cardboard with perimeter 14 inches, three parallel and equally spaced creases are made. The cardboard is then folded along the creases to make a rectangul

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: On a rectangular piece of cardboard with perimeter 14 inches, three parallel and equally spaced creases are made. The cardboard is then folded along the creases to make a rectangul      Log On


   

Question 1141797: On a rectangular piece of cardboard with perimeter 14 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 josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The length of the sides perpendicular to the 3 creases
is +4x+
Let then perimeter = +P+
+P+=+2%2A%284x%29+%2B+14+-+2%2A%284x%29+
+P+=+2%2A%284x%29+%2B+2%2A%28+7+-+4x+%29+
Now I can say what volume +V+ is
when folded along the creases
+V+=+x%5E2%2A%28+7+-+4x+%29+
+V+=+7x%5E2+-+4x%5E3+
Here's the plot:
+graph%28+500%2C+500%2C+-5%2C+5%2C+-7%2C+7%2C+7x%5E2+-+4x%5E3+%29+
It looks like +x+=+1.2+ might be close to max volume
+V+=+7%2A1.2%5E2+-+4%2A1.2%5E3+
+V+=+7%2A1.44+-+4%2A1.728+
+V+=+10.08+-6.912+
+V+=+3.168+
That looks pretty close to the graph
You can get closer with calculator
Get a 2nd opinion if needed