a square cardboard sheet is of 30cm length, four equal squares of equal dimensions are cut from the four corners of the sheet. The remaining sheet is folded to form an open box,what is the maximum volume of the box so formed?

• we all know the volume of cube is L*B*H
  now let dimension of corner cut is "X"
  then after folding the cardboard in order to make OPEN cube
  the dimension of OPEN cube will be (L-2x)*(B-2x)*(X) {for batter imagination you can do practically}.
  Now to maximize the value of product of (L-2x)*(B-2x)*(X).
  (d/dx)((L-2x)*(B-2x)*(X))=0
  now
  (d/dx)((30-2x)*(30-2x)*x)=0
  X^2 - 20X + 75=0
  then X=5,15 we can clearly say x!=15
  then X= 5
  now
  Volume is (30-10)*(30-10)*(5)
  =20*20*5
  =400*5
  2000 cm^3
• 9 years agoHelpfull: Yes(4) No(0)
• 2000 cm^3
  
  If square of side 'x' cm. is cut at each corner, then dimension of cuboid formed will have length=x, breadth=x and height=(30-x)/2
  Its volume=length*breadth*height=x*x*(30-x)/2=(30x^2-x^3)/2
  For maximum volume taking derivative and equating to Zero,
  so 60x-3x^2=0 or x=20
  Maximum volume with x=20 is (30*20^2 - 20^3)/2=2000 cm^3
• 9 years agoHelpfull: Yes(2) No(0)
• How height is 30-x by 2
• 9 years agoHelpfull: Yes(0) No(0)
• Let a square of x is cut from all the four corners and the box is formed.
  Now, the height of the box is x cm.
  And as from 30 cm two squares are cut so the length and breadth of the box will be (30-2x) cm.
  So, Volume of the box = length * breadth * height
  = (30-2x)*(30-2x)*x cm^3
  Now to find the maximum volume we need to make the derivative of volume to zero w.r.t variable x.
  so, after finding the derivative of volume we get X^2 - 20*X + 75 = 0.
  Now solving this quadratic equation we get two values of x which are 5 and 15.
  So x = 15 will give us 0 volume and x = 5 cm will give us volume = (30-10)^2*5 = 2000 cm^3.
  This is the maximum volume which we get.
  
• 9 years agoHelpfull: Yes(0) No(0)