SOLUTION: By cutting away identical squares from each corner of a rectangular piece of cardboard and folding up the resulting flaps, an open box may be made. (a) If the cardboard is 15 in.

Algebra ->  Test -> SOLUTION: By cutting away identical squares from each corner of a rectangular piece of cardboard and folding up the resulting flaps, an open box may be made. (a) If the cardboard is 15 in.      Log On


   



Question 298862: By cutting away identical squares from each corner of a rectangular piece of cardboard and folding up the resulting flaps, an open box may be made.
(a) If the cardboard is 15 in. long and 7 in. wide, find the dimensions of the box that will yield the maximum volume.

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
First write the equation for the volume of the box.
If you cut squares from each corner where the side of the square is x inches, then the volume of the resulting box when the flaps are folded up can be expressed as:
V+=+%2815-2x%29%287-2x%29X which simplifies to:
V+=+4x%5E3-44x%5E2%2B105x To find the value of x which yields the maximum of volume (V), take the first derivative of this equation and set it equal to zero.
dV%2Fdx+=+12x%5E2-88x%2B105 Set this equal to zero and solve for x;
12x%5E2-88x%2B105+=+0 Solve by the quadratic formula:x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a where a = 12, b = -88, and c = 105. You can do this and you should get:
x+=+5.833 or x+=+1.5
Discard x+=+5.833 because it would result in a negative value for the length of the bottom of the box: 7-2x+=+7-11.66=-4.66, so...
x+=+1.5inches.
The dimensions of the box are:
Length, L+=+15-2x=15-2%281.5%29+=+15-3=12inches.
Width, W+=+7-2x+7-2%281.5%29+=+7-3=4inches.
Height, h+=+x=1.5inches.
The maximum volume of the box would be:
V%5Bmax%5D+=+L%2AW%2Ah
V%5Bmax%5D+=+12%2A4%2A1.5
V%5Bmax%5D+=+72cubic inches.