00:01
In this problem, an open top box has to be constructed so that its base is twice as long as it is wide.
00:13
That means if its width is w, then its length is 2w.
00:18
And the volume of the box has to be 2 ,500 cm cube.
00:26
Here we denotes the volume.
00:28
We are asked to find the dimensions of this box so that requires minimum amount of cardboard for its making.
00:38
So here the amount of cardboard required is the total surface area a of the box.
00:46
So here let l denotes the length of the base of the box so that it is 2 w where w denotes the width of the box and let h denotes the height of the box.
00:59
Then the volume of the box is length times width times height so that volume 2 500 is 2 w times w h that is 2 500 is 2 w square h from this we get h is equal to 2 500 by 2 w square that is 1 250 by w square now the total surface area a of the boxes, the area of the base, since only the bottom side has to be considered since the box is open.
01:41
So the area of the base is the length of the base times its width plus there are two sides width width w and two sides corresponding to the length to w and they form the phases of the box.
01:59
So for the two sides with width h and height, width w and height h, the area is two times the width into height.
02:10
And those sites with length 2 w and height h, the area is twice the length that is 2 w times height.
02:20
So that the total surface area is 2w square plus 2wh plus 4 wx.
02:30
Now to get the area in terms of the width, substitute for height in terms of w so that this is 2 w square plus 2 w times 1 250 by w square plus 4 w times 1 ,050 by w square.
02:50
Now upon simplification we get the area as a function of the width w that is 2 w square plus 2 ,000 5 5 .5...