Question

From a square piece of cardboard paper of area size 9 m2 , squares of the same size are cut off from each corner of the paper. This allows for folding the paper to a box without a top. What values for the lengths of the squares only make sense? How do we have to choose the length of the squares to maximize the volume of the box? Verify that your answer is indeed a maximum.

Answer 1

Answer:

Max vol = 2 cubic metres

Step-by-step explanation:

Given that from a square piece of cardboard paper of area size 9 m2 , squares of the same size are cut off from each corner of the paper.

Side of the square = 3m

If squares are to be cut from the corners of the cardboard we have the dimensions of the box as

3-2x, 3-2x and x.

Hence x can never be greater than or equal to 1.5

V(x) = Volume = $V(x) = x(3-2x)^2\\= 9x+4x^3-12x^2$

We use derivative test to find the maxima

$V'(x) = 9+12x^2-24x\\V"(x) = 24x-24$

Equate I derivative to 0

$9+12x^2-24x=0\\x=1/2,3/2$

If x= 3/2 box will have 0 volume

So this is ignored

V"(1/2) <0

So maximum when x =1/2

Maximum volume

=$9(1/2)+4(1/2)^3-12(1/2)^2\\=2$ cubic metres