The dimension is 16 cm x 10 cm

Question

An open rectangular box is to be constructed by cutting square corners out of a piece of cardboard whose dimensions are given below and folding up the flaps. What dimensions will yield a box of maximum volume? Find the maximum volume

Answer 1

Beginning with a cardboard piece of dimension 16x10, we'll cut a square out of each corner. Since we don't know the dimensions of the square to cut out of the corner, we'll just assign it a side length of x. Once the sides are folded up, the volume of the box is length*width*height, where the length is (16-2x), the width is (10-2x), and the height is x. V = (16-2x)(10-2x)x. We can see here that the volume will be equal to zero at x = 0, 5, and 8. Expand the expression for the volume, and since we're looking for the max volume, it will either happen at some local maximum between these values. A value of x >= to 8 and < 0 is not useful, as that means we have a side of the box with a negative length. Take the derivative of the Volume with respect to x set it equal to zero (local max condition) and solve for the value of x that satisfy. disregard negative values of x.

Answer 2

So Brix Denzell A.,

Use the variable "x" to stand for the side length of each removed flap. the remaining box "base" is then 16-2x (because BOTH end flaps are gone!), and the width of the remaining box base is 10-2x, for the same reason.

Now when you fold the flaps up, how tall is the box? x for each flap, right? So the volume is length * width * height = (16-2x)*(10-2x)*x . The last time I checked, you could take the first derivative of that expression (multiply it out, then take the derivative term by term) -- and since there was a maximum of the original function, the first derivative is 0 at the value for x which maximizes the volume (as it always is at a maximum or a minimum). So you can solve the resultant equation, it's a quadratic.

--Cheers, --mr. d.