SOLUTION: Sharon has a sheet of cardboard 16 cm long and 10cm wide. She cut four identical squares from the four corners and then folded the sides of the remaining shape up to form an open b

Algebra ->  Volume -> SOLUTION: Sharon has a sheet of cardboard 16 cm long and 10cm wide. She cut four identical squares from the four corners and then folded the sides of the remaining shape up to form an open b      Log On


   



Question 1051129: Sharon has a sheet of cardboard 16 cm long and 10cm wide. She cut four identical squares from the four corners and then folded the sides of the remaining shape up to form an open box. What is hte maximum volume of the box that Sharon can make from her cardboard?
Found 2 solutions by josmiceli, ankor@dixie-net.com:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let the sides of the 4 squares = +x+
When the squares are cut out, and sides are
folded up, the sides of the box are:
+x+ cm
+10+-+2x+ cm
+16+-+2x+ cm
---------------------
The volume is:
+V+=+x%2A%28+10+-+2x+%29%2A%28+16+-+2x+%29+
+V+=+x%2A%28+160+-+32x+-+20x+%2B+4x%5E2+%29+
+V+=+4x%5E3+-+52x%5E2+%2B+160x+
+V+=+x%5E3+-+13x%5E2+%2B+40x+
Is this for a calculus course?
If so, Set the derivative = zero
+V%5B1%5D+=+3x%5E2+-+26x+%2B+40+
+3x%5E2+-+26x+%2B+40+=+0+
Use the quadratic formula
+x+=+-b+%2B-+sqrt%28+b%5E2+-+4%2Aa%2Ac+%29%29+%2F+%28+2%2Aa+%29+
+a+=+3+
+b+=+-26+
+c+=+40+
Plug in these numbers and find +x+
then plug +x+ back into +V+ to get
the maximum volume

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Sharon has a sheet of cardboard 16 cm long and 10cm wide.
She cut four identical squares from the four corners and then folded the sides of the remaining shape up to form an open box.
What is the maximum volume of the box that Sharon can make from her cardboard?
:
let x = the side of the square cut out
then the length and width of the box will be (16-2x) by (10-2x), the height = x
The volume equation, length * width * height
v = (16-2x)*(10-2x)*x
Foil
V = (160 - 32x - 20x + 4x^2)*x
V = x(4x^2 - 52x + 160)
V = 4x^3 - 52x^2 + 160x
The easiest way to find the max vol is by graphing this equation
+graph%28+300%2C+200%2C+-2%2C+6%2C+-10%2C+180%2C+4x%5E3-52x%5E2%2B160x%29+
you can see max vol occurs when x=2, which is 144 cu/in